Alekhya

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 53 Problem 1 Answer

Given: Number of shares = x  Price per share=$y

Annual dividend per share=$d

To find: The percent yield

Solution: Write the yield as a fraction.

Then convert the fraction to a decimal. Finally write the decimal as a percent.

Yield=Annual dividend per share Current price of one share

=d/y                                                [in fraction]

=d/y×100%                               [in percentage]

If we bought x shares of a stock for $y  per share and the annual dividend per share is $d then yield is 100 d/y%

Page 53 Problem 2 Answer

Given: For one share of Skroy Corporation:

Annual Dividend=$1.55

Today’s close=$x

Net change=+0.40

To find: The yield at yesterday’s close

Solution: First, find the closing of the

Yesterday’s close=Today’s close−Net change

=x−0.40

Yield=Dividend

Yesterday′ close

=1.55/x−0.40          [in fraction]

=1.55/x−0.40×100%    [in percentage]

When one share of Skro y Corporation stock pays an annual dividend of$1.55 and today’s Skro y closed at x dollars with a net change of +0.40, then the yield at yesterday’s close will be 100×1.55/x−0.40%

Page 54 Problem 3 Answer

Given: A corporation was paying a $2.10 annual dividend. The stock underwent an a3−for−2 split

To find: The new annual dividend per share

Solution: New annual dividend per share=b/a× Pre Split annual dividend per share

=2/3×2.10

=$1.40 (rounded off to two decimals)

A corporation was paying a$2.10 annual dividend and the stock underwent a a3−for−2 split so the post-split annual dividend per share is $1.40

Page 53 Problem 4 Answer

Given: Details of bond held by Adam Principal amount(P)=$1000

Rate of interest (R)=5.7% per year i.e.0.057

To find Total interest received if the time period (T) of holding bonds is 11years.

Solution: Interest=PRT

=1000×0.057×11

=627

If Adam holds the bond from Example 6 for 11 years, he will receive $627 in total interest.

Page 55 Problem 5 Answer

It is given: I believe non-dividend stocks aren’t much more than baseball cards.

They are worth what you can convince someone to pay for them.

The analogy of non-dividend stocks and baseball cards represents the similarity of two things in which value is determined by the demand, so they can be worthless in one moment and in another very expensive.

If we talk in terms of baseball cards, their value might vary a lot.

The card of some barely known baseball player is worth very little. It has only its intrinsic value.

However, if he becomes very famous, the value of the card increases drastically.

Some card-lovers are ready to pay huge amounts just to get that particular card.

And if people buy cards for their pleasure, that is completely fine.

But, if some person buys the card, or the stock share, with the hope to sell later for a higher price, we can notice the similarity.

Gaining profit from both the investments is dependent on the random value that the buyers will decide to give.

Page 55 Problem 6 Answer

Given: Years ago, Home Depot had an annual dividend of $0.90.

If you owned 4,000 shares of Home Depot. To find how much was received annually in dividends.

We will use the formula  Income dividends =Number of shares × Dividend p

Now the number of shares is given to be 4000 Dividend per share is $0.90

Hence Income dividends = Number of shares × Dividend per share

​=4000×0.90

=3600

​The amount that was received annually in dividends was $3,600.

Page 55 Problem 7 Answer

Given: you own r shares of a stock with an annual dividend of p dollars.

To express the  amount of your quarterly dividends algebraically.

We will use the formula:  Income dividends quarterly = Number of shares × Dividend per share quarterly

We have the number of shares as r.

Dividend per share annually is $p

Hence we get:​

Income dividends quarterly = Number of shares × Dividend per share quarterly

​=r×p/4

=rp/4

The amount of the quarterly dividends can be expressed as rp/4  dollars.

Page 55 Problem 8 Answer

Given: Mike owned 3000 shares of Merck Corporation and received a quarterly dividend check for $1140.

To find the annual dividend for one share of Merck.

We will use the formula: Annual divided per share = Income dividend annually ÷ Number of shares

Now we have the number of shares as 3000

Income dividend quarterly is $1,140

Therefore we get;​

Income Dividend annually =4× Income Dividend quarterly

​=4×$1,140

=$4560

and hence Annual divided per share = Income dividend annually ÷ Number of shares

​=$4560÷3000

=$1.52

The annual dividend for one share of Merck was $1.52.

Page 55 Problem 9 Answer

Given: The Walt Disney Company paid a $0.35 annual dividend on a day it closed at a price of $33.86 per share.

To find the annual dividend for 500 shares.

We will use the formula  Income dividends = Number of shares × Dividend per share

Now we have the number of shares as 500

Dividend per share is given to be $0.35

Therefore we get:​

Income dividends = Number of shares × Dividend per share

​=500×0.35

=175

The annual dividend for 500 shares was $175.

Page 55 Problem 10 Answer

It is given that :

Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to find the quarterly dividend for500 shares.

We will use the fact that there are four quarters in a year, so the dividend per share annually is annually dividend per share divided by 4.

Firstly, we will find the dividend per share quarterly.

Dividend per share quarterly = Dividend per share annually /4

Dividend per share quarterly=$0.35/4

Dividend per share quarterly=$0.0875.

Now, we will find the quarterly dividend.​​

Income dividends quarterly = Number of shares × Dividend per share quarterly​

Income dividends quarterly =500×0.0875

Income dividends quarterly =43.75.​​

We get $43.75.

The quarterly dividend for 500 shares was$43.75.

Page 55 Problem 11 Answer

It is given that : Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to express yield as a fraction.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, Yield= Annual dividend per share

Current price of one share

Yield=0.35/33.86

Yield=35/3386.​

We get the yield as a fraction:=35/3386.

Page 55 Problem 12 Answer

Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to find the yield to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, Yield= Annual dividend per share

Current price of one share

Yield=0.35/33.86

Yield=35/3386

Yield≈0.010

Yield=1.0% .​

The yield to the nearest tenth of a percent is 1.0%.

Page 55 Problem 13 Answer

We are given : A spreadsheet in which price per share and annual dividend are given.

The Black Oyster Corporation is going out of business. All of the corporate assets are being sold.

The money raised will be split by the stockholders.

We have to find among stockholders, the common or preferred, who receive money first.

Among the common and preferred stockholders, the preferred stockholders will receive money first, as the name “preferred” indicates that they will be preferred first than any common stockholder.

Also, preferred stockholders receive their dividends before the common stockholders and they also receive a set amount of dividends, which doesn’t vary too often, while the common stockholders will only receive dividends when the board of directions issues the dividends.

Among stockholders, the common or preferred, money will be received first by the preferred stockholders.

Page 56 Exercise 1 Answer

We are given: A table which gives the last price and the annual dividend for 15 corporations.

We have to compute the yield for each corporation and round answers to the nearest tenth of a percent.

We will be finding the yield for each corporation.

3 M Co Yield= Annual dividend per share

Current price of one share

Yield=2.00/76.90

Yield≈0.026

Yield=2.6% .

Alcoa. Inc​Yield=0.68/41.57

Yield≈0.016

Yield=1.6% .

American Express Co​Yield =0.72/46.15

Yield≈0.016

Yield=1.6% .​

American International Group, Inc

Yield =0.80/34.91

Yield≈0.023

Yield=2.3% .

American International Group, Inc

Yield =0.80/34.91

Yield ≈0.023

Yield =2.3%.

AT &T

Yield =1.60/39.51

Yield≈0.040

Yield=4.0% .​

Bank of America Corp​​​

Yield =2.56/33.87​

Yield≈0.076

Yield=7.6% .​

Boeing Co Yield =1.60/82.13

Yield ≈0.019

Yield =1.9% .

Caterpillar, Inc.

Yield =1.44/83.19

Yield≈0.017

Yield=1.7% .

Chevron Corp.

Yield =2.60/100.42

Yield≈0.026

Yield=2.6% .

Citigroup, Inc.

Yield =1.28/21.60

Yield≈0.059

Yield=5.9% .​

Coca-Cola Co/The

Yield =1.52/57.44

Yield≈0.026

Yield=2.6% .​

El Du Pont de Nemours & Co

Yield =1.64/47.63

Yield≈0.034

Yield=3.4% .​

Exxon Mobil Corp

Yield =1.60/90.43

Yield≈0.018

Yield=1.8% .​

By computing the yield for each corporation, we get :

3 M Co:2.6%

Alcoa, Inc:1.6%

American Express Co:1.6%

American International Group, Inc:2.3%

AT & T:4.0%

Bank of America Corp:7.6%

Boeing Co:1.9%

Caterpillar, Inc.:1.7%

Chevron Corp:2.6%

Citigroup, Inc.:5.9%

Coca-Cola Co/The:2.6%

El Du Pont de Nemours & Co :3.4%

Exxon Mobil Corp:1.8%.

Page 56 Exercise 2 Answer

We are given : Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the yield that was on Friday and then round answers to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, yield on friday=dividend on friday price on friday

yield on friday=1.6/44

yield on friday=0.036

yield on friday=36% .​

On Friday, the yield was:36%.

Page 56 Exercise 3 Answer

We are given : Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the price at which Revreg closed on Thursday and round answers to the nearest tenth of a percent.

We will be using the formula: Thursday closing = Friday closing − net change.

The Thursday closing is given by : thursday closing = Friday closing − net change

thursday closing =$44−$0.35

thursday closing =$43.65.​

$43.65 is the price at which Revreg closed on Thursday.

Page 56 Exercise 4 Answer

We are given: Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the yield at Thursday’s close.

We will be using the formula :

Thursday closing = Friday closing − net change.

Firstly, we will find the Thursday closing.

Thursday closing = Friday closing − net change

Thursday closing =$44−$0.35

Thursday closing =$43.65.

Now, we will find the yield at Thursday’s close.

yield on Thursday =dividend on Thursday price on Thursday

yield on Thursday =1.6/43.65

yield on Thursday =0.0366

yield on Thursday =3.66%

yield on thursday =3.7% .

3.7% is the yield at Thursday’s close.

Page 56 Exercise 5 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35

Thursday’s net change=$1.22.

We have to find the price at which Revreg closed on Wednesday and round answers to the nearest tenth of a percent.

We will be using the formula: Wednesday closing = thursday closing −thursday′s net change.

Firstly, we will find the Thursday closing.

thursday closing = Friday closing −friday′s net change

thursday closing =$44−$0.35

thursday closing =$43.65 .

We will find the Wednesday closing .

wednesday closing = thursday closing − thursday′s net change

wednesday closing =43.65−1.22

wednesday closing=42.43 .​

$42.43 is the price at which Revreg closed on Wednesday.

Page 56 Exercise 6 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35

Dividend on Wednesday=$1.60.

We have to find the yield at Wednesday’s close .

Firstly, we will find the Thursday closing .

thursday closing = Friday closing −friday′s net change

thursday closing =$44−$0.35

thursday closing =$43.65 .

We will find the Wednesday closing .

wednesday closing = thursday closing − thursday′s net change

wednesday closing =$43.65−$1.22

wednesday closing=$42.43 .

​The yield at Wednesday’s close is given by :

yield on wednesday =dividend on wednesday price on wednesday yield on wednesday =1.6

42.3 yield on wednesday =0.0377

yield on wednesday =3.77%

yield on wednesday =3.8% .

3.8% was the yield at Wednesday’s close .

Page 56 Exercise 7 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35.

By looking at the yields for Wednesday, Thursday, and Friday, we get that they are decreasing .

We have to explain why this decrease is not “bad news” to the investor who owns stock in Revreg.

The yield falls because of rising share prices, not because of falling dividends per share. Since

the dividend level has not changed, this is not “bad news” for the investor who owns shares in Revreg.

You will receive the same dividend amount regardless of the drop in performance.

And, another reason could be that the stock is a growth stock as its prices rise sharply every day.

Remember, growth stocks are bought for capital gains rather than dividends- buy low and sell high.

The decrease is not “bad news” to the investor who owns stock in Revreg because the amount of dividends did not actually decrease and there is possibly a growth stock .

Page 56 Exercise 8 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85.

We have to find the yield that was on Wednesday and then round answers to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share .

Therefore, yield on Wednesday=dividend on Wednesday

price on Wednesday

yield on Wednesday=2/61

yield on Wednesday=0.0327

yield on Wednesday=3.27%

yield on Wednesday=3.3% .

​On Wednesday, the yield was :3.3%.

Page 56 Exercise 9 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85.

We have to find the price at which Zeescore closed on Tuesday .

We will be using the formula :

Tuesday closing = Wednesday closing − net change.

The Tuesday closing is given by :

Tuesday closing = Wednesday closing − net change

Tuesday closing=61−(−0.85)

Tuesday closing=61+0.85

Tuesday closing=$61.85 .

$61.85 is the price at which Zeescore closed on Tuesday .

Page 56 Exercise 10 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85 .

We have to find the yield that was on Tuesday and then round answers to the nearest tenth of a percent.

We will find the formula :Tuesday closing = Wednesday closing − net change.

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85 .

Now, we will find the yield at Tuesday’s close .

yield on Tuesday =dividend on Tuesday price on Tuesday

yield on Tuesday =2/61.85

yield on Tuesday =0.0323

yield on Tuesday=3.23%

yield on Tuesday=3.2% .

​3.2% is the yield at Tuesday’s close .

Page 56 Exercise 11 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85

Net change on Tuesday=−1.96.

We have to find the price at which Zeescore closed on Monday .

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85 .

Now, we will find the Monday closing .

Monday closing = Tuesday closing − net change

Monday closing =61.85−(−1.96)

Monday closing =61.85+1.96

Monday closing =$63.81 .

​$63.81 is the price at which Zeescore closed on Monday .

Page 56 Exercise 12 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85

Net change on Tuesday=−1.96.

We have to find the Monday’s yield .

We will be using the formula :Monday closing = Tuesday closing − net change.

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85.​

We will find the Monday closing .

Monday closing = Tuesday closing − net change

Monday closing =61.85−(−1.96)

Monday closing =61.85+1.96

Monday closing =$63.81.​

Now, we will find the Monday’s yield.

yield on monday = dividend on monday price on monday

yield on monday=2/63.81

yield on monday=0.0313

yield on monday=3.13%

yield on monday=3.1% .

​Monday’s yield is3.1% .

Page 56 Exercise 13 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85 .

By looking at the yields for Monday, Tuesday, and Wednesday we get that they are increasing.

We have explain this increase is not “good news” to the investor who owns stock in

The yield is increasing and the reason is the price decrease of the stock, not because the dividends are increasing.

As the dividends are not increasing, the yield increase is not necessarily ‘good news’ to the investor who owns Zeescore stock because he or she will still receive the same amount of dividends despite the increase in yield.

The increase is not “good news” to the investor who owns stock in Zeescore because there is no increase in the amount of dividends.

Page 56 Exercise 14 Answer

We are given : Cost of shares=$1000

Annual Interest=6.34% or 6.34/100

=0.0634.

Time(in years)=1 year.

We have to find how much will Sascha receive in annual interest.

We will find the interest after one year.

Interest= Cost of shares×Annual interest×Time

Interest=1000×0.00634×1

Interest=63.4% .

Sascha will receive 63.4% in annual interest.

Page 56 Exercise 15 Answer

We are given: Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634.

Time(in years)=14 years.

We have to find how much will Sascha receive in interest if she holds the bond for 14 years.

We will find the interest after 14 years.

Interest after14 years=14× interest after 1 year

Interest after 14 years=14×63.4

Interest after 14 years=88.76%.

Sascha will receive 88.76%  in interest if she holds the bond for 14 years.

Page 56 Exercise 16 Answer

We are given : Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634

Price of stock=$46

Annual dividend=$2.

We have to find the yield.

We will find the yield.

yield =dividend

price of stock yield=2/46

yield=0.04347

yield=4.347%

yield=4.4% .​

The yield is 4.4%.

Page 56 Exercise 17 Answer

We are given: Cost of shares=$1000

Annual Interest​=6.34 %

=0.0634

Price of stock=$46

Annual dividend=$2

Time( in years)=14 years.

We have to determine which is higher, the yield on the stock or the interest rate on the bond.

Firstly, we will find the yield and then the interest.

We will find the yield.

yield =dividend price of stock

yield=2/46

yield=0.04347

yield=4.347% .

By rounding off ,we get :yield=4.4%.

We will find the interest after 14 years.

Interest after 14 years=14× interest after 1 year

Interest : =14×63.4

=88.76%.​

The interest rate on the bond is higher than the yield on the stock as the interest rate is 88.76%.

Page 56 Exercise 18 Answer

We are given: Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634

Time(in years)=1,14 years.

Bond matures after one year when :

Amount received=initial amount+ interest of 1year

Amount received=1000+0.0634×1000×1

Amount received=1000+63.4

Amount received=$1063.4.

Bond matures after 14 years when :

Amount received=1000+14(0.0634)(1000)

Amount received=1000+14(0.0634)

Amount received=1000+887.6

Amount received=$1887.6.

Lewis Corp. pays to Sascha when the bond matures :

After1 year :$1063.4

After14 years:$1887.6 .

Page 56 Exercise 19 Answer

We are given: Annual dividend per share=$2.46

Selling price per share=$54.24.

Selling price per share underwent a 3-for-1 split.

We have to find the new price of one share after the split.

As, we are given3−for−1split, therefore, the selling price before split will be divided by 3.

Therefore,Selling price after split = Selling price before split ÷3 Selling price after split =$54.24÷3 Selling price after split =$18.08.​

The new price of one share after the split is$18.08.

Page 56 Exercise 20 Answer

We are given: Annual dividend per share=$2.46

Selling price per share=$54.24

Number of shares=100.

The number of shares underwent a 3-for-1 split.

We have to find the number of shares you own after the split.

As we are given3−for−1split, therefore, the number of shares is multiplied by 3.

Number of shares after split = Number of shares before split ×3

Number of shares after split=200×3

Number of shares after split =600.

​The number of shares you own after the split is600.

Page 56 Exercise 21 Answer

We are given : Annual dividend per share=$2.46

Selling price per share=$54.24.

The annual dividend per share underwent a 3-for-1 split.

We have to find the annual dividend per share after the As, the annual dividend per share underwent a3−for−1 split, therefore, the annual dividend is divided by3.

Annual dividend after split = Annual dividend before split ÷3

Annual dividend after split =$2.46÷3

Annual dividend after split =$0.82.

The annual dividend per share after the split is$0.82.

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 138 Problem 1 Answer

Given: Principal amount(P)dollars

Rate of interest(R) compounded annually. i.e.

Time(T) year

To find: Amount of interest

Solution: Interest

PRT dollars will earn dollars in one year at a rate of compounded annually

Page 139 Problem 2 Answer

Given: p=4000

r=5% compounded semi-annually

=0.05

To find: Balance after one year

Solution: Accounts that pay interest semiannually have the interest added on twice each year (every six months).

We will first find the balance after six monthsThen calculate interest on that amount

For the first six months,

t=6 months  =0.5 years

I=PRT

=4000×0.05×0.5

=100

Now we will add the interest to the principal,

New principal=4000+100

=$4,100

Now we will compute interest for the next six months

I=PRT

=4100×0.05×0.5

=102.5

Now, we will add the interest to find the balance after one year

Balance after one year=4100+102.5

=4,202.5

Alex deposits$4,000 in a savings account that pays 5% interest compounded semiannually, then the balance after one year will be$4,202.5

Page 139 Problem 3 Answer

Given: p=$3000

r=4% compounded quarterly i.e.0.04

To find: Earnings in six months

Solution: Interest will be calculated every three monthsWe will find interest for the period of first three months and then again for the next three months

For the first three months

I=prt

=3000×0.04×0.25     [Convert 3 months in years]

=30

New principal=3000+30

=$3030

For the next three months

I=prt

=3030×0.04×0.25

=30.3

Total Interest=30+30.3

=$60.3

For the next three months

I=prt

=3030×0.04×0.25

=30.3

Total Interest=30+30.3

=$60.3

$3,000 earns$60.3 in six months at an interest rate of 4% compounded quarterly

Page 139 Problem 4 Answer

Given;  x dollars earn in one day at an interest rate of 5 % compounded daily

To find; Express the answer algebraically.

Accounts that pay interest daily earn interest every day.There are 365 days in a year.

r=5

r=0.05

​t=1 day

t=1 day =1/365 years

simple interest I=prt

I=1,000×0.05×1/365

≈0.13 interest to the principal. 1,000+0.13=1,000.13

The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.

This larger principal is used to compute the next day’s interest

Hence we conclude that The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.

This larger principal is used to compute the next day’s interest

Page 140 Problem 5 Answer

Given that On January 7, Joelle opened savings account with 900. It earned 3 % interest compounded daily.

On January 8, she deposited her first paycheck of76.22.

To do: find what was her balance at the end of the day on January 8.

From the given data the starting principal amount is 900 $ and given that at an interest of 3 %

it is compounded daily then we get 900(0.03)(1/365) deposited a paycheck of 76.22 $ on adding up all of them we get

900+900(0.03)(1/365)(76.22)

=976.29 $ is the balance

Therefore, the total balance at the end of the day is 976.29 $

Page 141 Problem 6 Answer

We have to find that how might those words apply to what you learned in this lesson

In the old days, a man who would save money was called a miser, because they just saved their money and didn’t earn anything with this money.

Thus the man wouldn’t profit from saving that money and also wouldn’t use this money for some other purpose.

However, nowadays, it is very unique and a kind of miracle to find someone who saves money, because it is not profitable to start saving money at an older age (as time is crucial in making savings grow) and thus you won’t gain much on your money (we do earn some due to compound interest).

Thus if you want to profit the most of your savings, then we need to start saving at a young age, but most people don’t tend to start saving at a young age and thus it is a kind of miracle to find someone who do as such.

A miser just saved their money and didn’t earn anything with this money.

However, nowadays someone who saved money is a wonder, because they would have to start at a young age if they want to profit from their savings.

Page 141 Problem 7 Answer

Given: Jerome deposits $3,700 in a certificate of the deposit that pays 6×1/2 interest, compounded annually.

We have to find that how much interest does Jerome earn in one year

Convert rate of interest to a decimal.

r=6×1/2%=6.5%=0.065

Use the simple interest formula I=prt

Substitute I=3760×0.065×1=244.4

Jerome will earn $244.4. as interest at the end of 1 year.

Page 141 Problem 8 Answer

Sally deposits $4,000 in a certificate of deposit that pays 6×3/4% simple interest.

What is her balance after one year

Convert rate of interest to a decimal.

r=6×3/4%=6.75%=0.0675

Use the simple interest formula I=prt

Substitute I=4000×0.0675×1=270

Add interest to the principal. 4,000+270=4,270

Sally’s balance after one year will be$4,270.

Page 141 Problem 9 Answer

Given: Pierre deposits $9,000 in a certificate of deposit that pays 8% interest, compounded semiannually.

To find: How much interest does the account earn in the first six months?

What is the balance after six months

Convert rate of interest to a decimal. r=8%=0.08

Convert 6 months to years.t=6 months =0.5 years

Use the simple interest formula ​I=prt

By Substituting

I=9000×0.08×0.5=360

Pierre’s account will earn $360 as interest after 6 months.

The balance after 6 months will be $9,360​

Add interest to the principal. 9,000+360=9,360

The balance after 6 months will be $9,360

Page 141 Problem 10 Answer

Given: Kevin has x dollars in an account that pays 2.2% interest, compounded quarterly.

We have to Express his balance after one quarter algebraically.

Convert rate of interest to a decimal. r=2.2%=0.022

Step 2 Convert 3 month to years.

t=3 months =1/4 years =0.25 years Use the simple interest formula

By Substituting

I=x×0.022×0.25=0.0055x

Kevin’s balance after one quarter will be $1.0055x.​

Add interest to the principal. x+0.0055x=1.0055x

Kevin’s balance after one quarter will be $1.0055x.

Page 141 Exercise 1 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to Find the first quarter’s interest.

p=3500

r=7.5%=0.075

t=1 quarter =0.25yrs first quarter interest ∣=prt=35000.0750.25=65.625

We get, $65.625

Page 141 Exercise 2 Answer

Given: Liam deposits $3,500  in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the first quarter’s ending balance.

Given: Principal:$3,500

Result part (a): The first quarter interest is $65.625

First quarter interest =$65.625≈865.63

The first quarter’s ending balance if the principal increased by the first quarter interest:

First quarter’s ending balance = Principal + First quarter interest

​=$3,500+$65.63

=$3,565.63

​We get,

$3,565.63

Page 141 Exercise 3 Answer

Given: Liam deposits $3,500

in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the second quarter’s interest.

Given:

p= Principal =$3,565.63 (First quarter’s ending balance found in part (b))r= Interest rate =7×1/2 %=7.5%=0.075

t= Time in year=1

Quarter =1/4 years =0.25years

The interest is the product of the principal, interest rate and time expressed in years.

I=prt=3,565.63×0.075×0.25≈$66.86

We get,

$66.86

Page 141 Exercise 4 Answer

Given: Liam deposits $3,500

in a saving account that pays7x1/2% interest, compounded quarterly.

We have to find the second quarter’s ending balance.

Given: Principal=$3,565.63 (First quarter’s ending balance found in part (b))

Result part (c): The second quarter interest is $66.86

Second quarter interest =66.86

The second quarter’s ending balance is the principal increased by the second quarter interest:

Second quarter’s ending balance = Principal + Second quarter interest

=$3,565.63+$66.86

=$3,632.49

​We get,

$3,632.49

Page 141 Exercise 5 Answer

Given: Liam deposits $3,500  in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the third quarter’s interest.

Given: p= Principal =$3,632.49 (First quarter’s ending balance found in part (d)) r= Interest rate =7×1/2%=7.5%=0.075

t= Time in year=1 Quarter=1/4 years =0.25 years

The interest is the product of the principal, interest rate and time expressed in years.

I= prt =3,632.49×0.075×0.25≈$68.11

we get,

$68.11

Page 141 Exercise 6 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the third quarter’s ending balance.

Given: Principal =$3,632.49 (First quarter’s ending balance found in part (d))

Result part (e): The third quarter interest is $68.11

Third quarter interest =68.11

The third quarter’s ending balance is the principal increased by the third quarter interest:

Third quarter’s ending balance = Principal + Second quarter interest

​=$3,632.49+868.11

=$3,700.60

​The third quarter’s ending balance is : $3,700.60

Page 141 Exercise 7 Answer

Given: Liam deposits $3,500

in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the fourth quarter’s interest.

Given:p= Principal r= (First quarter’s ending balance found in part (f))r= Interest rate =7×1/2%=7.5%=0.075

t= Time in year =1 Quarter=1/4 years =0.25 years

The interest is the product of the principal, interest rate and time expressed in years.

I=prt=3,700.60×0.075×0.25≈$69.39

We get,

$69.39

Page 141 Exercise 8 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

To find: What is the balance at the end of one year?

Given: Principal =$3,700.60 (First quarter’s ending balance found in part (f))

Result part (g): The fourth quarter interest is $69.39

Fourth quarter interest =69.39

The fourth quarter’s ending balance is the principal increased by the fourth quarter interest:

Fourth quarter’s ending balance = Principal + Second quarter interest

​=$3,700.60+$69.39

=$3,769.99

​The balance at the end of one year is : $5,769.99

Page 141 Exercise 9 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find that how much interest does the account earn in the first year?

Total interest in 1 st year = Balance after 1 year- original principal amount

=3769.98-3500=269.98

The amount earned in first year is : $269.98

Page 141 Exercise 10 Answer

Given : Janine opens a savings account with a deposit of $720.

The account pays3.4 percent interest, compounded daily.

We have to find first day’s interest.

We will use above formula .

Here, First convert rate of interest into decimal:r=3.4%=0.034​

Convert 1 day to years:t=1day =1/365 years​

Now, use simple interest formula: I=720×0.034×1/365≈0.07

Hence, the First day’s interest is approximately 7 cents.

Therefore, First day’s interest is approximately 7 cents.

Page 142 Exercise 11 Answer

We have given a table,

We have to find the missing amounts in the table.

We will find by using above information.

No principle amount initially thus

(1) Opening balance = 0

(1) Deposit on 10 August = $4,550

(3) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=0+4550−0

=4550

(4) The interest is the product of the principle, interest rate and time in years:

I=prt

=4550×0.04×1/365

≈0.50​

Thus interest rate is $0.5

(5)  Ending Balance = principal+interest

Ending Balance =4550+0.5=4550.50

ending balance is $4,550.50

(6)  Next day opening = previous day ending balance

Opening bal aug 11=4550.50

(7)  Deposit on August 11=300

(8)  principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=4550.50+300−0

=4850.50

​(9) The interest is the product of the principle, interest rate and time in years:

I=prt

=4850.50×0.04×1/365

≈0.53interest rate is $0.53

(10)  Ending Balance = principal+interest

Ending balance =4850.5+0.53=4851.03

(11)  Next day opening = previous day ending balance

Opening bal aug 12=4851.03

(12)  withdrawal=900

(13)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle= Opening balance + Deposit − Withdrawal

=4851.03+0−900

=3951.03

(14)  The interest is the product of the principle, interest rate and time in years:

I=prt

=3951.03×0.04×1/365

≈0.43

​(15)  Ending Balance = principal+interest

Ending Bal=3951.03+0.43=3951.46

The answers of the given blanks in the table are given above. All the answers are in USD.

Page 142 Exercise 12 Answer

Given: On December 18 of a leap year, Stacy opened a savings account by depositing $6,000.

The account pays 3.45% interest, compounded daily.

On December 19 she deposited $500, and on December 20 she withdrew $2,500.

By using this data we will fill all the blanks in the table.

(1)  no principal amount initially

Thus opening balance is = 0

(2) Deposit = $6000

(3)  Withdrawal =0

(4) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=0+6000−0

=6000

​(5)  The interest is the product of the principle, interest rate and time in years:

I=prt

=6000×0.0345×1/365

≈0.57

​(6)  Ending Balance = principal+interest

Ending Balance =6000+0.57=6000.57

(7)  Next day opening = previous day ending balance

Opening bal =6000.57

(8)  Deposit =500

(9)  Withdraw =0

(10)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=6,000.57+500−0

=6,500.57

(11)  The interest is the product of the principle, interest rate and time in years:

I=prt

=6500.57×0.0345×1/365

≈0.61

​(12) Ending balance =6500.57+0.61=6501.18

(13) Next day opening = previous day ending balance

Opening bal =6501.18

(14) deposit =0

(15)  withdrawal=2500

(16) ​ Principle = Opening balance + Deposit − Withdrawal

=6,501.18+0−2,500

=4,001.18

(17)  The interest is the product of the principle, interest rate and time in years:

I=prt

=4,001.18×0.0345×1/365

≈0.38

​(18)  Ending Balance = principal+interest

Ending Bal=4001.18+0.38=4001.56

The answers of the given blanks in the table are given above. All the answers are in USD.

Page 142 Exercise 13 Answer

Given : On May 29, Rocky had an opening balance of x dollars in an account that pays 3% interest, compounded daily. He deposits y dollars.

We have to express his ending balance on May30 algebraically.

We will use above formula .

Here, Consider that the deposit occurs on May 23

So, p=Principle

=Opening balance+Deposit

=$(x+y)

r=Interest rate

=3%

=0.03

t=Time in years

=1day

=1/365years (as there are 365 days in a year)

The interest is the product of the principle, interest rate and time in years:I=prt

=(x+y)×0.03×1/365

=x×3/100×1/365

=3(x+y)/36,500

The ending balance is then the principle increased by the interest: Ending balance

=p+I

=x+3(x+y)/36,500

=36,500x+3x+3y/36,500

=36,503x+3y/36,500

Now, p=Principle=ClosingbalanceMay29

t=$(36,503x+3y/36,500)

r=Interest rate

=3%

=0.03

The interest is the product of the principle, interest rate and time in years:I=prt

=(36,503x+3y/36,500)×0.03×1/365

=(36,503x+3y/36,500)×3/100×1/365

=(36,503x+3y/36,500)×3/36,500

=3(36,503x)+3(3y)/36,5002

=109,509x+9y/1,332,250,000​

The ending balance is then the principle increased by the interest:Ending balance =p+I

=36,503x+3y/36,500+109,509x+9y/1,332,250,000

=36,500(36,503x)+36,500(3y)/1,332,250,000+109,509x+9y/1,332,250,000

=1,332,359,500x+109,500y/1,332,250,000+109,509x+9y/1,332,250,000

=1,332,359,500x+109,500y+109,509x+9y/1,332,250,000

=1,332,469,009x+109,509y/1,332,250,000

Hence, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.

Therefore, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.

Page 142 Exercise 14 Answer

Given :Linda has d dollars in an account that pays 3.4% interest, compounded weekly. She withdraws w dollars.

We have to express her first week’s interest algebraically.

We will use above formula .

Here, p=Principle

=Opening balance−Withdrawal

=d−w​

r=Interest rate

=3.4%

=0.034

t=Timeinyears

=1week

=1 /52 years (as there are 52 weeks in a year)

Now, The interest is the product of the principle, interest rate and time in years: ​I

=prt

=(d−w)×0.034×1/52

=(d−w)×34/1000×1/52

=34(d−w)/52,000

=17(d−w)/26,000.

Hence, The first week’s interest is 17(d−w)/26,000.

Therefore, The first week’s interest is17(d−w)/26,000

Page 142 Exercise 15 Answer

Given: The table represents the compound interest calculations for an account that pays 2% interest compounded daily. Represent a–g algebraically.

Opening balance = P

Deposit   = D

Withdrawal = $0

By using this data we will fill the table .

1) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=P+D−0

=P+D

(2)  The interest is the product of the principle, interest rate and time in years:

I=prt

=(P+D)×0.02×1/365

=(P+D)×2/100×1/365

=2(P+D)/36,500

=P+D/18,250

(3)  The ending balance is then the principle increased by the interest:

Ending balance =p+I

=(P+D)+(P+D)/18,250

=(P+D)(1+1/18,250)

=(P+D)(18,250/18,250+1/18,250)

=(P+D)(18,251/18,250)

=18,251(P+D)/18,250

(4)  The opening balance on February 3 is the ending balance on the previous day and thus is the ending balance on February 2.

Opening balance on February 3= Opening balance on February 2

=18,251(P+D)/18,250

(5)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=18,251(P+D)/18,250+0−W

=18,251(P+D)/18,250−W

(6)  The interest is the product of the principle, interest rate and time in years:

I=prt

=(18,251(P+D)/18,250−W)×0.02×1/365

=18,251(P+D)−18,250W/18,250×2/100×1/365

=18,251(P+D)−18,250W/18,250×1/18,250

=18,251(P+D)−18,250W/18,2502

=18,251(P+D)−18,250W/333,062,500​

(7)  The ending balance is then the principle increased by the interest:

Ending balance =p+I

=18,251(P+D)/18,250−W+18,251(P+D)−18,250W/333,062,500

=18,251(P+D)−18,250W/18,250+18,251(P+D)−18,250W/333,062,500

=18,250(18,251)(P+D)−333,062,500W/333,062,500+18,251(P+D)−18,250W/333,062,500

=333,080,750(P+D)−333,062,500W/333,062,500/+18,251(P+D)−18,250W/333,062,500

=333,080,750(P+D)−333,062,500W+18,251(P+D)−18,250W/333,062,500

=333,099,001(P+D)−333,080,750W/333,062,500

Therefore,

1)The principle used to compute interest is the new balance is P+D

2)Interest = =P+D/18,250

3) Ending Balance =18,251(P+D)/18,250

4)Opening Balance =18,251(P+D)/18,250

5)The principle used to compute interest =18,251(P+D)/18,250−W

6)The interest is the product of the principle, interest rate and time in years:  =18,251(P+D)−18,250W/333,062,500

7)Ending Balance:- 333,099,001(P+D)−333,080,750W/333,062,500

Page 142 Exercise 16 Answer

Given: George had an opening balance of m dollars in an account that pays 2.25% interest compounded monthly.

On the last day of the month, he made a deposit equal to twice his opening balance.

We have tp express his ending balance on the last day of the month algebraically.

We will use above formula .

Here, Opening balance =m

Deposit =2m (twice the opening balance)

p  =Principle

= Opening balance + Deposit

=m+2m

=3m

​r= Interest rate

=2.25%

=0.0225

​t=Time in years

=1month

=1/12 years  (as there are 12 months in a year)

Now,The interest is the product of the principle, interest rate and time in years: ​I

=prt

=3m×0.0225×1/12

=3m×225/10000×1/12

=3m×9/400×1/12

=(3m)(9)/400(12)

=27m/4,800​

The ending balance is then the principle increased by the interest:Ending balance

=p+I

=3m+27m/4,800

=14,400m/4,800+27m/4,800

=14,400m+27m/4,800

=14,427m/4,800

=4,809m/1,600​

Hence,

The ending balance on the last day of the month is 4,809m /1,600.

Therefore,The ending balance on the last day of the month is 4,809m/1,600.

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 42 Problem 1 Answer

Given: Rob purchased stock and he sold it at a loss.

To find: Express his net proceeds algebraically.

We will find the difference in the stock purchased and the stock sold.

For purchasing the total money spent will be p+40 and when the stock is sold, we get the net money by removing brokerage fee as h−1%h=0.99h.

Hence, we will get the net proceeds as 0.99h−(p+40)=0.99h−p−40.

Thus, we can say that when Rob purchased stock and he sold it at a loss then his net proceeds algebraically will be 0.99h−p−40.

Page 43 Problem 2 Answer

Given:

To find: How do those words apply to an investor? How do those words apply to a stockbroker?

We will use the statement as per the definition with reference of the stock.

Since we know that sometimes the price of stock increases or decreases rapidly.

When we have purchases a stock but it’s value start decreasing then it will be our loss.

However, as we know that we are the pilot.

So, we can say that we will decide when we want to sell the stock and when we want to purchase more.

Thus, we can say that we will decide when we want to sell the stock and when we want to purchase more.

Page 43 Problem 3 Answer

Given that The ticker shows trades of stock in Hewlett-Packard (HPQ), Exxon- Mobil (XOM), and Chevron (CVX).

HPQ 6K47.29 ▼ 0.23 XOM 3K92.67 ▲ 0.08 CVX 9K100.38 ▼ 0.22 We have to find how many shares of Hewlett-Packard were sold

Given that Hewlett-Packard (HPQ) in the given ticket as HPQ6K47.29▼0.23 As we know that the number of shares sold will be the number oresent right side of the name of the company.

So on observing the given ticket HPQ6K47.29▼0.23

We can say that the number of shares sold as 6K  that is nothing but 6000.

There fire the number of shares sold are 6000.

Page 43 Problem 4 Answer

Given that The ticker shows trades of stock in Hewlett-Packard (HPQ), Exxon- Mobil (XOM), and Chevron (CVX).

HPQ 6K47.29 ▼ 0.23 XOM 3K92.67 ▲ 0.08 CVX 9K100.38 ▼ 0.22

We have to find what was the total value of all the HPQ shares sold

Given that Hewlett-Packard (HPQ) in the given ticket as HPQ6K47.29▼0.23

As we know that the number of shares sold will be the number present right side of the name of the company.

We can say that the number of shares sold as 6000 And we know the price per share is the number present directly to the left side after the name of the company.

That is  Price per share =47.29

Now we have the total value of the shares sold as the product of the number of shares and the price value of each share that is

Total value sale=Number of shares sold×Price per share

=6000×47.29

=283740​

Hence we have the total value of all the HPQ shares sold as 283740

Page 43 Problem 5 Answer

Given that The ticker shows trades of stock in Hewlett-Packard (HPQ), Exxon- Mobil (XOM), and Chevron (CVX).

HPQ 6K47.29 ▼ 0.23 XOM 3K92.67 ▲ 0.08 CVX 9K100.38 ▼ 0.22

Also given that broker charged her 1% commission.

We have to find the total cost of her investment.

Given that Hewlett-Packard (HPQ) in the given ticket as HPQ6K47.29▼0.23

As we know that the number of shares sold will be the number present right side of the name of the company.

We can say that the number of shares sold as 6000 And we know the price per share is the number present directly to the left side after the name of the company.

That is  Price per share =47.29

Now we have the total value of the shares sold as the product of the number of shares and the price value of each share that is

Total value sale = Number of shares sold × Price per share

​=6000×47.29

=283740​

Now we have the comission as Commission

=1%× Total value sale

=0.01×283740

=2837.40​

Now we have that the total cost of investment as the sum of total value and the comission.

Total cost of the investment

= Total value sale + Commission

=283740+2837.40

=286577.40

Hence we have the total cost of her investment as 286577.40

Page 43 Problem 6 Answer

Given that The ticker shows trades of stock in Hewlett-Packard (HPQ), Exxon- Mobil (XOM), and Chevron (CVX).

HPQ 6K47.29 ▼ 0.23 XOM 3K92.67 ▲ 0.08 CVX 9K100.38 ▼ 0.22 Also given that broker charged her 1.5% commission.

We have to find how much money did the broker receive and also need to round to the nearest cent.

Given that Exxon- Mobil (XOM)  ticket as XOM3K92.67▲0.08

As we know that the number of shares sold will be the number present right side of the name of the company.

We can say that the number of shares sold as 3000 And we know the price per share is the number present directly to the left side after the name of the company.

That is  Price per share =92.67

Now we have the total value of the shares sold as the product of the number of shares and the price value of each share that is Total value sale

= Number of shares sold × Price per share

=3000×92.67

=278010

​Now we have the comission as Commission

=1.5%× Total value sale

=0.015×278010

=4170.15​

Hence we have the commission value as $4170.15

Page 43 Problem 7 Answer

Given that The ticker shows trades of stock in Hewlett-Packard (HPQ), Exxon- Mobil (XOM), and Chevron (CVX).

HPQ 6K47.29 ▼ 0.23 XOM 3K92.67 ▲ 0.08 CVX 9K100.38 ▼ 0.22

Also given that her discount broker, who charges $28 per transaction.

We have to how much money did Lisa receive from the above sale after the broker took his fee

Given that Chevron (CVX) ticket as CVX9K100.38▼0.22

As we know that the number of shares sold will be the number present right side of the name of the company.

We can say that the number of shares sold as 9000 And we know the price per share is the number present directly to the left side after the name of the company.

That is  Price per share =100.38

Now we have the total value of the shares sold as the product of the number of shares and the price value of each share that is

Total value sale=Number of shares sold×Price per share

​=9000×100.38

=903420

 

Now we have the  the profit of the sale is the total price value subtracted with the broker fees that is Profit of the sale

= Total value sale − Broker fee

=903420−28

=903392

The amount of money Lisa receive from the above sale after the broker took his fee is 903392

Page 44 Problem 8 Answer

Given: Lenny bought x  shares of stock for $y per share last month. He paid his broker a flat fee of $20 He sold the stock this month for $p per share, and paid his broker a 2% commission.

To express Lenny’s net proceeds algebraically.We will get the purchase price and the selling price to get the net proceeds.

purchase price of x shares = xy

total purchase  cost = xy+20

Selling price of the shares will be= xp

Total selling cost = xp−0.02xp

Hence it is given by: 0.98xp

Therefore the net proceeds will be:

Net proceeds = total selling cost – total purchase  cost

= 0.98 x p-(xy+20)

=x(0.98p-y)-20

Lenny’s net proceeds is given by: x(0.98p−y)−20

Page 44 Problem 9 Answer

Given:  Darlene purchases $20,000 worth of stock on her broker’s advice and pays her broker a 1.5 % broker fee.

She sells her stock when it increases to $28,600 two years later, and uses a discount broker who  charges $21per trade.

To compute Darlene’s net proceeds after the broker fees are taken out.We will get the purchase price and the selling price to get the net proceeds.

The purchase price = Price of shares bought + the broker fees.

Hence the total purchase will be 20,000

Selling price of shares =28600

Total selling cost will be ⇒28600−21=28579

Net proceeds = total selling cost – total purchase  cost

Net proceeds ⇒28579−20300=8279

Darlene’s net proceeds after the broker fees are taken out is $8279.

Page 44 Problem 10 Answer

Given: Ron bought x dollars worth of stock and paid a y percent commission.

Dave purchased p dollars worth of stock and paid a q percent commission, where x>p.

To pick numbers for x,y,p,q such that  Ron’s commission is less than Dave’s. We will get the purchase price for both.

Purchase price of shares for Ron is given to be x.

Hence commission Ron paid is: (y/100)x=xy/100

Purchase price of shares for Dave is given to be y.

Hence commission that Dave paid will be (q/100)p=pq/100

Therefore for Ron’s commission to be less than Dave we have:​

xy/100<pq/100

xy<pq  since  q/y>x/p>1​

For Ron’s commission to be less than Dave we must have: xy<pq

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 36 Problem 1 Answer

Given: Brett used money he received as a gift for high school graduation.

To find gross capital gain or loss. Product of shares and price of each stock will be total purchasing power.

Since we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will be $4000 and total selling price will be$2,433.

Hence, the gross capital will be 2433−4000=−1567<0.

Thus, we can say that $1567 was his capital loss when Brett used money he received as a gift for high school graduation.

Page 36 Problem 2 Answer

Given: Kelvin bought 125 shares of stock for 68.24 per share.

To find: What was his capital gain? Since we know that product of shares and price of each stock will be total purchasing power.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will be125×68.24=8530 andtotal selling price will be125×85.89=10,736.25.

So, we can say that capital gain will be 10,736.25−8530=2206.25.

Thus, we can say that when Kelvin bought 125 shares of stock for 68.24 per share then his capital gain will be $2206.25.

Page 38 Problem 3 Answer

Given: Zach bought 200 shares of Goshen stock years ago for $21.35per share.

He sold all 200 shares today for$43 per share.

To find: gross capital gain. Product of shares and price of each stock will be total purchasing power.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will be 200×21.35=4270 and total selling price will be 200×43=8600.

Hence, the gross capital will be $8600−$4270=$4330.

Thus, we can say that $4330 will be the gross capital gain for Zach.

Page 38 Problem 4 Answer

Given: Mitchell bought 600 shares of Centerco two years ago for 34.50 per share.

To find: What was the percent increase in the price per share.We will use the formula of price increase as the difference in price divided by old price given.

Since, we know that the price increase will be the difference in price divided by old price given as

Percent increase

= New price − Old price / Old price

=38.64−34.50/34.50

=4.14/34.50

=0.12

=12%

​Thus, we can say that the percent increase in the price per share will be 12% from the shares bought.

Page 38 Problem 5 Answer

Given: Mitchell bought 600 shares of Center co two years ago for 34.50 per share.

To find: was the total purchase price for the 600 shares.We will use the unitary method which will be taken as for one unit of reference.

Since, we know that price of a single share will be $34.50.

Thus, using the unitary method, we will get total purchase price as 600⋅34.50=20,700.

Thus, we can say that $20,700 will be the total purchase price for the 600 shares from the shares bought.

Page 38 Problem 6 Answer

Given: Mitchell bought 600  shares of Centerco two years ago for34.50 per share.

To find the total selling price for the 600 shares.We will use the unitary method which will be taken as for one unit of reference.

Since, we know that price of a single share will be$38.64.

Thus, using the unitary method, we will get total purchase price as 600⋆38.64=$23184.

Thus, we can say that$23,184 will be the total selling price for the 600 shares from the shares bought.

Page 38 Problem 7 Answer

Given: Mitchell bought 600 shares of Centerco two years ago for 34.50 per share.

To find percent capital gain for the 600 shares. Total gain will be the difference between the selling price and the purchase price.

Total gain will be the difference between the selling price and the purchase price given as 23184−20700=$2484 which will be equivalent to2484∗100/23184=12%.

Thus, we can say that 12% was the percent capital gain for the 600 shares from the shares bought.

Page 38 Problem 8 Answer

Given: Mitchell bought 600 shares of Centerco two years ago for 34.50 per share.

To find: the percent increase in the price of one share compare to the percent capital gain for all shares.

We know that percentage in gain will not be dependent on the number of shares bought.

Since, we have seen that the percent increase in the price of one share is 12%.

As, percentage in gain will not be dependent on the number of shares bought.

Thus, we will be same values.

Thus, we can say that when the percent increase in the price of one share compared to the percent capital gain for all 600 shares we get the same value as 12%.

Page 38 Problem 9 Answer

Given: Tori bought x shares of Mattel stock for m dollars per share.

To find: Express her capital gain or loss algebraically. Product of shares and price of each stock will be total purchasing power.

Since, we know that total purchase price for x shares will be m x and selling price will be x y.

Hence, we will get the capital gain as yx−mx=(y−m)⋅x

Thus, we can say that capital gain or loss algebraically will be (y−m)⋅x for the Tori.

Page 38 Problem 10 Answer

Given: Ramon bought x shares of Xerox stock for a total of 40,000.

To find: Express the price he paid per share algebraically.

We will use the unitary method which will be taken as for one unit of reference.

Using the unitary method, we can say that when the price of x shares is 40,000

then we get the price of one share as 40,000/x.

Thus, we can say that the price Ramon paid per share algebraically will be 40,000/x.

Page 38 Exercise 1 Answer

Given: Joe bought 200 shares in the Nikon corporation for12.25 per share.

To find Joe’s capital gain. Product of shares and price of each stock will be total purchasing power.

Since we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will be 200×12.25=2450 and total selling price will be200×$31.27=$6254

Hence, the gross capital will be$6254−$2450=$3804.

Thus, we can say that $3804will be the gross capital gain for Joe.

Page 38 Exercise 2 Answer

Given: Joe bought 200  shares in the Nikon corporation for 12.25 per share.

To find: Joe’s capital gain as a percent, rounded to the nearest percent.

We will use the formula of price increase as the difference in price divided by old price given.

Since, we know that the price increase will be the difference in price divided by old price given as

Percent increase= Gross capital gain

Total purchasing price=3804/2450

=1902/1225

≈1.553

=155.3%​

Thus, we can say that Joe’s capital gain as a percent, rounded to the nearest percent will be 155.3%.

Page 38 Exercise 3 Answer

Given: If you bought and then sold 300 shares at these prices.

To find: what was your loss.Product of shares and price of each stock will be total purchasing power.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will be 32×300=9600 and total selling price will be20×300=6000.

Thus, we get the loss will be 9600−6000=3600.

Thus, we can say that loss will be$3600 when we bought and then sold 300 shares at these prices.

Page 38 Exercise 4 Answer

Given: General Motors stock fell from 32 per share in 2006 to 20 per share during 2008.

To find: Express your loss as a percent of the purchase price. Product of shares and price of each stock will be total purchasing power.

Since, we know that the price increase will be the difference in price divided by old price given as

(3600∗100)/9600=37.5%

Thus, we can say that 37.5% will be loss as a percent of the purchase price.

Page 38 Exercise 5 Answer

Given: Elliott purchased shares of Microsoft for 28 per share.

To find: At what price will he sell his shares? We will use the unitary method which will be taken as for one unit of reference.

Since we know that the price increase will be the difference in price divided by old price given as1.2∗28=33.6.

Thus, we can say that at $33.6 will Elliott sell his shares when he purchased shares of Microsoft.

Page 39 Exercise 6 Answer

Given: a share of stock in the Coca-Cola Company sold for 39.

To find: Express the increase in price as a percent of the price.

We will use the formula of price increase as the difference in price divided by old price given.

Since, we know that the price increase will be the difference in price divided by old price given as

Percent increase= New price − Old price / Old price

=56−39/39

=17/39

≈0.4359

=43.59%​

Thus, we can say that the percent increase in the price per share will be 43.59% for Coca-Cola Company.

Page 39 Exercise 7 Answer

Given: Alexa purchased 700  shares of Campagna Corporation stock for x dollars per share.

To find: Did Alexa have a gross capital gain or a gross capital loss.We will use the relation to compare the values to determine the gain or loss.

When we have the relation as y<x where y is the selling value and xis the purchase value.

So, we can say that

​y⋅700<x⋅700

⇒700y−700x<0 which is a loss.

Thus, we can say that Alexa have a gross capital loss for 700 shares of Campagna Corporation stock.

Page 38 Exercise 8 Answer

Given: Alexa purchased 700 shares of Campagna Corporation stock for x dollars per share.

To find: Who is correct, Alexa or Tom.We will use the relation to compare the values to determine the gain or loss.

From the values that is 700y−700x/700x, we can write it as

​700y−700x/700x

=700(y−x)/700x

=y−x/x which gives us the same value given by Tom.

Thus, we can say that both Alexa and Tom are correct as they gives the same value algebraically for the percent of the loss.

Page 39 Exercise 9 Answer

Given: Zeke bought g shares of stock for w dollars per share.

To find: Express the total purchase price of all the shares algebraically.We will use the unitary method which will be taken as for one unit of reference.

Using the unitary method we can say that when g shares of stock for w dollars per share then we get the total purchase price of all the shares algebraically as g⋆w=gw.

Thus, we can say that the total purchase price of all the shares algebraically will be gw for Zeke.

Page 39 Exercise 10 Answer

Given: Zeke bought g shares of stock for w dollars per share.

To find:  Express the capital gain algebraically.Since, we know that product of shares and price of each stock will be total purchasing power.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will beg w,and

total selling price will be

g×w+40/100×g×w

=g×w+0.4×g×w

​Hence, the gross capital will be(g×w+0.4×g×w)−g×w=0.4×g×w.

Thus, we can say that the capital gain algebraically will be 0.4×g×w for Zeke.

Page 39 Exercise 11 Answer

Given: Zeke decides to sell his shares. To find: Express the total selling price of all the shares algebraically.

We will use the unitary method which will be taken as for one unit of reference.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total purchasing price will beg w and total selling price will be

g×w+40/100×g×w

=1×g×w+0.4×g×w

=(1+0.4)×g×w

=1.4×g×w

Thus, we can say that 1.4×g×w will be the total selling price of all the shares algebraically for Zeke.

Page 39 Exercise 11 Answer

Given: Jake bought d shares of stock for x dollars per share years ago.

To find: Represent half of the d shares algebraically.We will be using the algebraic method to get the result.

Using the algebraic method, we can say that the value of half of the shares will be d/2=0.5d.

Thus, we can say that half of the shares will be 0.5dfor Jake.

Page 39 Exercise 12 Answer

Given: Jake bought d shares of stock for x dollars per share years ago.

To find: Represent the capital gain earned on each of the shares that were sold algebraically.

we know that product of shares and price of each stock will be total purchasing power.

Since, we know that product of shares and price of each stock will be total purchasing power.

Thus, we can say that total selling price will be

140=100(yd−xd)/xd

⇒y/x

=2.4

⇒y=2.4x

Hence, the gross capital will be

y−x/x×100​

=100⋆(2.4x−x)/x=100∗1.4

=$140

Thus, we can say that capital gain earned on each of the shares that were sold algebraically will be $140 for Jake.

Page 39 Exercise 13 Answer

Given: Jake bought d shares of stock for x dollars per share years ago.

To find: Represent the capital gain earned on all of the shares that were sold algebraically.

We know that product of shares and price of each stock will be total purchasing power.

Since, we know that total selling price will be

​140=100(yd−xd)/xd

⇒y/x

=2.4

⇒y=2.4x

​Thus, we get the capital gain earned on all of the shares as

​​=100∗((y−x)d/2)/(xd/2)

=100∗(2.4−1)x/x

=1.4100

​=140%​

Thus, we can say that the capital gain earned on all of the shares that were sold algebraically will be 140%for Jake.

Page 39 Exercise 14 Answer

Given: Jake bought a share of stock for x dollars per share years ago.

To find: Represent the total value of the shares that were sold algebraically.

We will use the unitary method which will be taken as for one unit of reference.

From the selling value as y=2.4x, we get the value of total share hold as

yd/2=2.4xd/2

=1.2xd

Thus, we can say that the total value of the shares that were sold algebraically will be 1.2xd

for Jake.

Page 39 Exercise 15 Answer

Given: Jake bought d shares of stock for x dollars per share years ago.

To find: did investing in d shares result .in a capital gain or loss for Jake?

We will use the unitary method which will be taken as for one unit of reference.

Since, we know that purchase value is xd and selling value is 1.2 xd that gives us the net gain value as

1.2xd−xd=0.2xd>0

Thus, we can say that investing in the d shares result in a capital gain for Jake.

Page 39 Exercise 16 Answer

Given: Ahmad sold 125 shares of stock for x dollars.

To find: How much did he originally pay for the shares of stock We will use the unitary method which will be taken as for one unit of reference.

Using the unitary method we can say that when 125 shares of stock are purchased for $32.75  per share we will get original pay for the shares of stock as

125⋅32.75=4093.75$

Thus, we can say that $4093.75

Ahmad originally pay for the shares of stock.

Page 39 Exercise 17 Answer

Given: Ahmad sold 125 shares of stock for x  dollars.

To find: Write an inequality that represents an amount such that Ahmad made money from the sale of the stocks.

We will try get the maximum profit and will adjust the result accordingly.

Let us take the price of a selling stock as y.

Thus, in order to make the profit, such that Ahmad made money from the sale of the stocks we will write y>32.75 which end up in giving the gain.

Thus, we can say that y>32.75 is the inequality that represents an amount such that Ahmad made money from the sale of the stocks.

Page 39 Exercise 18 Answer

Given: Suppose Ahmad lost money on the stocks. To find: Write an inequality that represents an amount such that Ahmad lost no more than 1,000  from the sale of the stocks.

We will try get the maximum profit and will adjust the result accordingly.

Since we know that the selling price of stocks will be 125x and the purchase price as 4093.75.

So, in order to maintain the loss, we take

​4093.75−125x≤1000

⇒x≥24.75​

Thus, we can say that an inequality that represents an amount such that Ahmad lost no more than 1,000 from the sale of the stocks will be x≥24.75.

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market

Financial Algebra 1st Edition Chapter 3: Banking Services

Page 132 Problem 1 Answer

Given -5.51%,51/2%,55/8%,5.099%,5.6%

We need to arrange the given interest rates in descending order.

We have :

5.51%,51/2%,55/8%,5.099%,5.6%

These are equivalent to :

5.51%,5.50%,5.625%,5.099%,5.6%

Arranging these in descending order, we get :

5.625%,5.6%,5.51%,5.50%,,5.099%

We obtain :

5.625%,5.6%,5.51%,5.50%,,5.099%

Page 133 Problem 2 Answer

Given – Mae has $891  in her account. A$7 fee is charged each month the balance is below $750.

She withdraws$315. She makes no deposits or withdrawals for the next x months.

We need to express her balance algebraically.

New Balance= Prior Balance−Reduction​

=891−315

=576

New Balance=Prior Balance− Reduction​

=576−7

=$569.00​

The new balance is: $569.00

Page 133 Problem 3 Answer

We will find the simple interest that is earned on $4,000 in 31/2 years at an interest rate of 5.2% We can calculate the simple interest as follows :

I =p⋅r⋅t

i=(4000)(0.052)(3.5)

i=812

The simple interest is $812.

Page 133 Problem 4 Answer

We will find the simple interest that $800 earn in 300 days in a non-leap year at an interest rate of 5.71%.

We will round to the nearest cent.

Here,t=300/365 years

i=p⋅r⋅t

i=800⋅0.0571⋅300/365

i=$37.55​

The simple interest is: $37.55

Page 134 Problem 5 Answer

Given – A two-year simple interest account pays 31/4% interest to earn $300 in interest.

We will find the principal.We will use the formulai=p⋅r⋅t.

We have :

i=p⋅r⋅t

300=p⋅3.25⋅2

∴p=46.1538

​The principal amount is$46.1538.

Page 134 Problem 6 Answer

We will determine how long will it take $10,000 to double at 11% simple interest.

We will use the formula:i=p⋅r⋅t

For the given data, we get :

i=p⋅r⋅t

20000=10000⋅0.11⋅t

t=18.1818​

The time is : 18.1818 years

Page 134 Problem 7 Answer

Marcos deposited $500 into a 2.5-year simple interest account.

He wants to earn$200 interest. We will find interest rate that must the account pay.

We will use the formula :i=p⋅r⋅t.

We know :

i=p⋅r⋅t

From the given data, we get :

200=500⋅2.5⋅r 2⋅10 5⋅25

=r

r=0.16

We get r=16%

We get the rate as 16%.

Page 135 Problem 8 Answer

We will arrange the following interest rates in ascending order :

3.4%,3.039%,33/16%,3.499%,31/2%

We will first convert all the rate of interests in decimal form.

1.​3.4%=3.400

2.​3.039%=3.039  ( no need to make changes)

3.​33/16=(16×3+3)/16→3.188

4.​3.499=3.499

5.​31/2=(3×2+1)/2→3.500

Comparing the decimal forms, we get,

3.039<3.188<3.400<3.499<3.500

Hence,3.039<33/16%<3.4%<3.499%<31/2%

We conclude :

3.039<33/16%<3.4%<3.499%<31/2%

Page 135 Problem 9 Answer

Given – Josh has a savings account at a bank that charges a $10 fee for every month his balance falls below$1,000.

His account has a balance of$1,203.44 and he withdraws $300.

We will determine his balance be in six months if he makes no deposits or withdrawals.

We will use the known facts.

The new balance is the previous balance decreased by the withdrawal.

$1,203.44−$300=$903.44

We note that the balance is below $1,000 and thus Josh has to pay a charge of$10

per month on each of the six months (when this balance is on the account).

Total fee: $10×6=$60

The balance after 6 months is then the previous balance decreased by the total fee of $60.

$903.44−$60=$843.44

His balance in six months if he makes no deposits or withdrawals is$843.44.

Page 135 Problem 10 Answer

Given – John, Paul, and George are having a disagreement over interest rates.

John says that 63/4% can be expressed as 6.75%.

George thinks that 63/4% can be expressed as 0.0675.

Paul remembers converting percents to equivalent decimals and thinks it can be expressed as0.0675%.

We will determine who is correct and who is incorrect.

John and George are correct. Paul’s conversion is incorrect.

63/4%=6.75%

=0.0675.

​These representations are correct.

While converting percentage to decimal form, the percentage sign needs to be dropped.

Paul converted it into decimal, yet kept the percent sign.

That was his mistake!

John and George are correct. Paul’s conversion is incorrect.

Page 135 Problem 11 Answer

Given – Beth and Mark would like to put some savings in the bank.

They most likely will not need this money for4

years, so Beth wants to put it in a four-year CD.

Mark wants to put the money in a passbook savings account.

We will tell the advantage and disadvantages of a CD.

Beth and Mark would like to put some savings in the bank.

They most likely will not need this money for 4 years, so Beth wants to put it in a 4−year CD.

Mark wants to put the money in a savings account.

The disadvantage of Cd are that Cd has a penalties if money is withdrawn before maturity, and the advantage of CD is that Cd have highest interest rate.

Page 135 Problem 12 Answer

We will find the simple interest on a $2,350 principal deposited for six years at a rate of 4.77%.

Principal amount(p)=$2350

No of years/ time(t)=6

Rate of interest(r)=4.77%

=0.0477

Simple Interest (I)=ptr

=2350×6×0.0477

=$672.57

​The simple interest is$672.57.

Page 135 Exercise 1 Answer

Given – Ryan depos its $775 in an account that pays 4.24%

simple interest for four years. Brian deposits $775 in an account that pays4.24%

simple interest for one year.We will determine the Ryan’s interest after the four years.

Principal(p)=$775

Rate of interest

(r)=4.24%

=0.0424

​No of years/time(t)=4 years

Simple interest=p t r

=775×4×0.0424

=131.44 dollars

Hence simple interest after 4 years is$131.44.

The simple interest after 4 years is$131.44.

Page 135 Exercise 2 Answer

Ryan deposits $775 in an account that pays 4.24% simple interest for four years.

Brian deposits $775 in an account that pays 4.24% simple interest for one year.We will find Ryan’s balance after four years.

Principal(p)=$775

Rate of interest

(r)=4.24%

=0.0424

​No of years/time(t)=4 years

Simple interest=p t r

=775×4×0.0424

=131.44 dollars

Hence simple interest after 4 years is$131.44

Amount = simple interest +principal

=775+131.44

=906.44  dollars

Hence amount after 4 years is$906.44

The amount after 4years is $906.44.

Page 135 Exercise 3 Answer

Ryan deposits $775  in an account that pays 4.24% simple interest for four years.

Brian deposits$775  in an account that pays 4.24%  simple interest for one year.

We will determine the interest that Ryan’s account earn the first year.

Principal(p)=$775

Rate of interest

(r)=4.24%

=0.0424​

No of years/time(t)=1 year

Simple interest =p t r

=775×1×0.0424

=32.86 dollars

Hence simple interest after 1 year is$32.86.

The simple interest after 1 year is$32.86.

Page 135 Exercise 4 Answer

Ryan deposits $775 in an account that pays 4.24% simple interest for four years.

Brian deposits$775 in an account that pays 4.24% simple interest for one year.

We will determine the interest that Ryan’s account earn the fourth year.

We will use the known facts.

We note that : i=p⋅r⋅t

From the given data, we get :

i=775⋅0.0424⋅4

i=131.44

​The simple interest is$131.44.

Page 135 Exercise 5 Answer

Ryan deposits $775  in an account that pays 4.24% simple interest for four years.

Brian deposits $775 in an account that pays 4.24% simple interest for one year.

We will determine  Brian’s interest after the first year.

Principal(p)=$775

Rate of interest

(r)=4.24%

=0.0424

​No of years/time (t)=1 years

Simple interest=p t r

=775×1×0.0424

​=32.86 dollars

Hence simple interest after 1 years is$32.86.

The simple interest after 1 years is$32.86.

Page 135 Exercise 6 Answer

Ryan deposits $775  in an account that pays 4.24% simple interest for four years.

Brian deposits $775 in an account that pays 4.24% simple interest for one year.

We will determine Brian’s balance after the first year.

Principal(p)=$775

Rate of interest

(r)=4.24%

=0.0424​

No of years/time (t)=1 years

Simple interest=p t r

=775×1×0.0424

=32.86 dollars

Hence simple interest after 1 years is$32.86

Amount= simple interest+ principal =32.86+775

=807.86 dollars

Hence amount after the first year is$807.86

The amount after the first year is$807.86.

Page 135 Exercise 7 Answer

Ryan deposits $775 in an account that pays 4.24% simple interest for four years.

Brian deposits $775  in an account that pays 4.24% simple interest for one year.

Here, Brian withdraws all of the principal and interest after the first year and deposits it into another one-year account at the same rate.

We will determine his interest for the second year rounded to the nearest cent.

In part (f), we determines that the balance after the first year is$807.86.

We then put the $807.86 in a one-year account at the same rate:

p=Principle =$807.86

r= Interest rate =4.24%

=0.0424

​Time t=1 year

The simple interest is the product of the principal, interest rate and time expressed in years.

=p r t

=807.86×0.0424×1≈$34.25

​Thus, the interest during the second year is $34.25.

The interest during the second year is $34.25.

Page 135 Exercise 8 Answer

Given – Ryan deposits $775  in an account that pays 4.24% simple interest for four years.

Brian deposits $775 in an account that pays 4.24% simple interest for one year.

We will compare the interest Brian earns with the interest Ryan earns for the second year.

We will tell who earned more interest.

We know, Simple interest=p t r

If we compare the values of p,t&r, for Ryan and Brian, we get that all the three values are equal. Hence the product p t r will also be equal for both.

Hence Ryan and Brian earn the same interest.

We conclude that Ryan and Brian earn the same interest.

Page 136 Exercise 9 Answer

Given: Principal =$2,560

Interest Rate =51/8%

Time =17 months

​To find the simple interest.We will use the formula Simple Interest = Principal ⋅ Rate of interest. Time.

Substituting the given values in simple interest formula as I=p r t

=2560×0.05125×17/12

≈$185.87

The simple interest for​

Principle =$2,560

Interest rate =51/8%

Time =17 months is $185.87.

Page 136 Exercise 10 Answer

Given:Interest =$450

​Principle =$450

Interest rate =14%

​To find the time when$450 can be doubled for the given interest rate.We will use the simple interest formula as t=I/pr.

Substituting the given values in the formula t=I/pr, we get

450/450×0.14

≈7.1429

It takes$450 approximately 7.1429 years to double at a simple interest rate of 14%.

Page 136 Exercise 11 Answer

Given:Interest =$450

Principle =$450

Interest rate =100%

​​To find the time such that the given interest gets doubled in the given interest rate.We will use the formula t=I/pr.

Substituting the given values in the simple interest formula as t=I/pr

=450/450×1

=1

It takes$450 one year to double at a simple interest rate of 100%.

Page 136 Exercise 12 Answer

Given:​​

Interest =$900

Principle =$9,500

​Time =19 months

​To find the rate of interest so that we can earn the given amount of interest.

We will use the formula r=I/pt.

Substituting the given values in the simple interest formula as r=I/pt ​

=900/9500×19/12

=108/1805≈0.0598

=5.98%

​The interest rate needed for$ 9,500 to earn $900 in 19 months is 5.98%.

Page 136 Exercise 13 Answer

Given: Assume $20,000 is deposited into a savings account.

Bed ford Bank offers an annual rate of 4% simple interest for five years.

Slick Bank offers a rate of 20% simple interest for one year.

To find that which earns more interest.

We will find out the simple interest for both banks using the formulaI=prt and compare them.

For Bedford Bank, we are given

Principle =$20,000

Interest rate =4%

Time =5 years

​so for them the simple interest is

​I=p r t

I=20000×0.04×5

I=4000

​For Slick Bank, we are given

Principle =$20,000

Interest rate =20%

Time =1 year

so for them the simple interest is

I=prt

I=20000×0.2×1

I=4000

Therefore, we get that the interest in both the cases is same and it is$4000.

Interest for both the banks is same and it is$4000.

Page 136 Exercise 14 Answer

Given: A couple is planning a savings account for a newborn baby.

They start with$3,450 received in newborn baby gifts.

If no deposits or withdrawals are made, to find the balance of the account if it earns simple interest at 5% interest for18

years. We will use the formula I=p r t.

We are given

Principle =$3,450

Interest rate =5%

Time =18 years

So using these values, we get the simple interest is

I=p r t

I=3,450×0.05×18

I=3,105

​If no deposits or withdrawals are made, the balance of the account if it earns simple interest at 5% interest for18 years is$3,105.

Page 136 Exercise 15 Answer

Given: Ron estimates that it will cost $400,000 to send his daughter to a private college in18 years.

He currently has $90,000 to deposit in an account.

To find that what simple interest rate must his account have to reach a balance of $400,000 in 18 years.

We will use the formula r=I/pt.

We are given that

Interest =$400,000−$90,000

Interest =$310,000

Principle =$90,000

Time =18 years

using these values the interest rate is r=I/pt

r=310000

90000×18≈0.19

r=19%

​The simple interest rate must his account have to reach a balance of $400,000 In 18 years is 19%.

Page 136 Exercise 16 Answer

Given:

For A₂ to compute the interest.We will use the formulaI=prt.

From the given table, we get that for row one, we have

Interest = A₂

Principal =B₂

Interest rate =C₂

Time =D₂​

So the simple interest for these values using the formulaI=prt becomes

A₂=B₂×C₂×D₂.

In a spreadsheet, this can be written as A₂=B₂⋅C₂⋅D₂.

The simple interest for A₂ in the table is A₂=B₂⋅C₂⋅D_2.

Page 136 Exercise 17 Answer

Given:

For B2 to compute the principal. We will use the formula I=p r t.

We are given that

Interest = A₂

Principal =B₂

Interest Rate =C₂

Time =D₂

​using these values we get that the principal formula is p=I/rt

B₂=A₂

C₂×D₂​

In a spreadsheet, we write it as B₂=A₂/(C₂⋅D₂)

The value of the principal for B₂ in the table is B₂=A₂/(C₂⋅D₂).

Page 136 Exercise 18 Answer

Given:

For C2 to compute the interest rate.We will use the formulaI=prt.

We are given that

Interest=A₂

Principal l =B₂

Interest Rate =C₂

Time =D₂

so using these values, the formula for interest rate is

r=I/pt

C₂=A₂

B₂×D₂

C₂=A₂/(B₂⋅D₂)​

The formula of interest rate for C₂ in the table is C₂=A₂/(B₂⋅D₂).

Page 136 Exercise 19 Answer

Given:

ForD₂ to compute time in years, given the interest, rate, and principal. We will use the formula I=p r t.

We are given that

Interest=A₂

Principal =B₂

Interest Rate =C₂

Time =D₂​ using these values the formula for time in years is

t=I/pr

D₂=A₂

B₂×C₂

In a spreadsheet, we write it as D₂=A₂/(B₂⋅C₂).

The formula of time in years for D₂ in the table is D₂=A₂/(B₂⋅C₂).

Page 136 Exercise 20 Answer

Given: Zoe creates a spreadsheet to make simple interest calculations.

The user inputs values for the principal, rate, and time in years in row 2.

Write each formula.

To find: For E₂ compute the time in months, given the time in years.

All values are entered or calculated in row 2.

By taking into account the column tables in row 1, we then note:

Time in years=D₂

Time in months=E₂

Since there are 12 months in a year, the time in month is 12 times the time in years:

E₂=12×D₂

In a spreadsheet, a product is most often entered using x instead of ×, and a division is entered using.

E2=1₂xD₂

We get, E₂=12⋅D₂

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 125 Problem 1 Answer

Given: We have given a bank statement and monthly cycles of checks.

To find: Name some reason why the check has not been clear.

Many people and businesses hold onto checks and do not deposit or cash them immediately.

Reasons a Bank Balance Will Differ from a Company’s Balance

Some of the reasons for a difference between the balance on the bank statement and the balance on the books include:

Outstanding checks

Deposits in transit

Bank service charges and check printing charges

Errors on the company’s books

Electronic charges and deposits that appear on the bank statement but are not yet recorded in the company’s records

If an outstanding check of the previous month does not clear the bank in the current month, the check will remain on the list of outstanding checks until the month that it does clear the bank.

In the bank reconciliation process, the total amount of the outstanding checks is deducted from the balance appearing on the bank statement.

Hence, we prove that why a check may not have cleared during the monthly cycle and appear on the bank statement.

Page 126 Problem 2 Answer

Given: outstanding checks.

To find: Write a formula to calculate the sum of the outstanding checks.

In the bank reconciliation process, the total amount of outstanding checks is subtracted from the ending balance on the bank statement when computing the adjusted balance per bank.

(No adjustment is needed to the company’s general ledger accounts, since the outstanding checks were recorded when they were issued.)

Hence, the formula to calculate the sum of the outstanding checks the total amount of outstanding checks is subtracted from the ending balance on the bank statement when computing the adjusted balance per bank.

Page 126 Problem 3 Answer

Given: Nancy has a balance of $1,078 in her check register. The balance on her bank account statement is $885.84.

Not reported on her bank statement are deposits of $575 and $250 and two checks for $195 and $437.84.

To find: Is her check register balanced? Explain.

Yes; 885.84 + 575 + 250 – 195 – 437.84 = 1078.

Since her revised statement balance equals her check register balance, the check register has been reconciled.

Hence, we prove that her revised statement balance equals her check register balance, the check register has been reconciled.

Page 127 Problem 4 Answer

Given: Ken filled out this information on the back of his bank statement.

To find: Find Ken’s revised statement balance. Does his account reconcile?

Revised Statement Balance = Ending Balance + Deposits – Checks Outstanding

$=197.10+600-615.15$

$=181.95$

which is not equal to check register balance hence the account does not reconcile

Hence, we found Ken’s revised statement balance is 181.95$ which is not equal to check register balance hence the account does not reconcile

Page 128 Problem 5 Answer

Given that Donna has a checking account that charges0.15

for each check written and a monthly service charge of 9.75.

We will write a formula that Donna can use each month to find the fees she will be charged. Identify any variable you use in the formula.

Let F be the total fees payable.

Let n  be the number of checks written

Fee per check=0.15dollar.

So, total fee of n

checks=0.15n

Also, service charge=9.75 dollars

Then, total fee = service charge+total check fee

So, F=9.75+0.15n

The required formula is :

F=9.75+0.15n

Page 130 Problem 6 Answer

We will fill in the missing balances in Raymond Marshall’s check register.

We will use the spreadsheet from www.cengage.com/school/math/financialalgebra to determine if Raymond’s checking account reconciles with his statement.

We note that all balance totals are missing in the check register and thus we have to calculate these balances for every row of the check register.

The new balance is the previous balance increased by the deposit or decreased by the payment.

We also add a check mark if the check mark is already present in the check register or if the payment/deposit is mentioned in the monthly statement.

In the check register, we note that there are five rows with no check mark and thus these five transactions were not included in the monthly statement.

Payment of$38.50 on 12/30

Payment of$100.00 on1/12

Payment of $85.00 on1/12

Payment of$80.00 on1/25

Deposit of$950.00 on2/1

The ending balance on the monthly statement is$1,378.57.

Ending balance=$1,378.57

The revised statement balance is then the ending balance increased by the outstanding deposits and decreased by the outstanding payments.

Revised statement balance= Ending balance+Outstanding deposits−Outstanding payments

​=$1,378.57+$950.00−$38.50−$100.00−$85.00−$80.00

=$2,025.07

We note that the revised statement balance of $2,025.07 is the same as the final balance total of$2,025.07  on the check register, thus the bank statement and check register reconcile.

The bank statement and check register reconcile.

We obtain :

653.30​583.30​544.80​516.80​396.80​340.07​240.07​155.07​1155.07​1,075.07​2,025.07

Page 127 Problem 7 Answer

Given: We have given a bank statement and monthly cycles of check.

To find: Name some reason why check has not been clear.

Gross income means income before expenses and tax etc.

Net income means income after expenses and tax etc. Everyone must reconcile their gross and net incomes every year.

Errol Lynn says he needs to reconcile his ‘gross’ habits, drinking, smoking, womanizing against his actual income, which of course would never be enough.

Hence, we conclude that Errol Lynn says he needs to reconcile his ‘gross’ habits, drinking, smoking, womanizing against his actual income, which of course would never be enough.

Page 127 Problem 8 Answer

Given: We have given a bank statement.

To find: write a formula and a statement for Hannah.

S=B+D-C

Revised Statement Balance = Ending Balance + Deposits – Checks Outstanding if S is equal to R then the account reconcile

Hence, the formula is Revised Statement Balance = Ending Balance + Deposits – Checks Outstanding if S is equal to R then the account reconcile

Page 127 Problem 9 Answer

Given: we have given a bank statement.

To find: What is the mean number of checks Jill wrote per month during the last four months?

Checks per month = total checks / no of months

Total checks. =14+19+23+24=80

No of months =4

So, checks written per month. =80 / 4=20 checks

Hence, the mean number of checks Jill wrote per month during the last four months is 20.

Page 127 Problem 10 Answer

Given: Given a check statement.

To find: estimate how much Jill expects to pay in per-check fees each month after she switches to the new account.

Mean checks per month. $=20 ( from Q. 5a) Fees per check $= 0.2

So fees estimated to be paid in per check fees each month = no of checks times fee per check 20\0.2$= 4

Hence, Jill expects to pay in per-check fees each month after she switches to the new account is $4.

Page 127 Problem 11 Answer

Given: We have given a check statement.

To find: Estimate the total monthly fees Jill will pay each month for the new checking account.

Monthly maintenance fees $= 12.50

Fees paid for checks written per month $= 4( Q.5B)

So, total fees paid. $=12.50+4=16.50 dollars

Hence, he total monthly fees Jill will pay each month for the new checking account is $16.50

Page 127 Problem 12 Answer

Given: Rona filled out this information on monthly statement.

To find: Find Rona’s revised statement balance. Does account reconcile?

Revised Statement Balance = Ending Balance + Deposits – Checks Outstanding

$=725.71+610.00-471.19$

$=864.52$ which is equal to check register balance hence the account reconcile

Hence, we found Rona’s revised statement balance is 864.52$ which is equal to check register balance hence the account reconcile

Page 130 Problem 13 Answer

When comparing his check register to his bank statement, Donté found that he had failed to record deposits of55.65,103.50,and25.00.

We will determine the total of these amounts and how will he use this information to reconcile his account.

We will use the known facts.

Deposits that were not recorded:$55.65,$103.50,$25.00

Let us first determine the total of all unrecorded deposits:55.65+103.50+25.00=184.15

Thus we note that the total of all unrecorded deposits is$184.15.

We will then reconcile his account by subtracting$184.15 from the ending balance and also noting this in the check register.

The total of these amounts is$184.15.

We will reconcile his account by subtracting$184.15 from the ending balance and also noting this in the check register.

Page 130 Problem 14 Answer

Given -d>c

Alisha has a February starting balance of $678.98 in her checking account.

During the month, she made deposits that totaled d dollars and wrote checks that totaled cdollars.

Here E=her ending balance on February28. We will write an inequality using E and the starting balance to show the relationship of her starting and ending balances for each condition.

If d>c, then that means that the February starting balance will increase because total deposits are greater than total withdrawals.

Thus,E should be greater than the February starting balance.

For example, letd=$100 and c=$50:

Ending balance

​=$678.98+$100−$50

=$728.98

​Ending balance is greater than starting balance.

We get E>$678.98

The required inequality is :

E>$678.98

Page 130 Problem 15 Answer

Given -d<c Alisha has a February starting balance of$678.98 in her checking account.

During the month, she made deposits that totaled d dollars and wrote checks that totaled d dollars. Here,E= her ending balance on February 28.

We will write an inequality using E and the starting balance to show the relationship of her starting and ending balances for each condition.

If d<c, then that means that the February starting balance will decrease because total deposits are lower than total withdrawals.

Thus,E should be lower than the February starting balance.

For example, let d=$50 and c=$100:

Ending balance

=$678.98+$50−$100

=$628.98

Ending balance is lower than starting balance.

Hence, E<$678.98

The required inequality is : E<$678.98

 

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 57 Problem 1 Answer

Let’s follow the corporation Ipath.B Sugar (SGGB) between August 13, 2018 and September12,2018.

Date	Open	High	Low	Close	Volume
Aug13	37.81	37.83	37.46	37.57	35,000
Aug 14	38.13	38.35	37.32	37.36	12,100
Aug 15	38.15	38.15	37.42	37.62	8,600
Aug 16	38.06	38.22	3.53	37.67	14,600
Aug 17	37.49	37.51	37.13	37.35	29,300
Aug 20	36.9	37.17	36.65	37.17	17,200
Aug 21	37.44	37.53	37.04	37.53	14,800
Aug 22	37.03	37.5	36.5	37.49	20,100
Aug 23	36.85	37.36	36.85	37.02	4,500
Aug 24	37.98	38.42	37.52	37.72	27,900
Aug 27	38.15	38.68	37.96	38.68	17,400
Aug 28	37.86	38.28	37.76	37.83	11,300
Aug 29	38.05	38.22	37.71	38.22	8,600
Aug 30	38.18	38.95	38.09	38.8	50,800
Aug 31	39.25	39.37	38.49	39.05	22,300
Sep 4	39.05	39.13	38.69	39.04	9,400
Sep 5	39.52	40.05	39.5	39.96	9,100
Sep 6	39.85	40.02	39.52	39.52	2,800
Sep 7	40.27	40.43	40.04	40.26	6,000
Sep 10	40.92	41.29	40.72	41.29	21,300
Sep 11	41.07	41.2	40.83	40.9	8,600
Sep 12	41.33	42.53	41.33	42.47	13,700

 

Now, we will write the WPO daily.

WPO Daily

The bar is red, if the opening price is larger than the closing price .

The bar is green, if the closing price is larger than the opening price,

The bar is bounded vertically by the opening price and the closing price, and all the bars need to have the same width.

The lowest point of the vertical line attached to the bottom of a bar represents the low price of the day.

The highest point of the vertical line attached to the top of a bar represents the high price of the day.

Volume

The height of the bars has to be equal to the volume, while the width of all bars has to be the same.

The bar is colored red if and only if the shares were mainly sold.

The bar is colored green if the shares were mainly bought.

We get the graph :

We have discussed the trend over the three-weeks and include any major corporate news that might have affected the trend and we get the graph :

Page 57 Problem 2 Answer

We have to discuss the possibility of purchasing shares of stock for a corporation you are interested in following .

Answer

When purchasing stocks, you must consider all relevant aspects, such as the company’s present financial/economic condition, the country’s financial/economic status, the financial/economic situation of the company’s major consumers, and so on.

This is significant since all of these factors have an impact on stock market pricing.

It is critical to purchase stocks at the correct time for the proper firm, namely when the stock price is low and the stock price is predicted to climb (much) in the near future.

Furthermore, it is critical to sell the stocks at the proper time in order to maximize your profit, specifically when the stock’s value is high and when the stock’s value is projected to decline (significantly) in the near future.

A stock market’s value can fluctuate dramatically over a short period of time, thus it’s critical to maintain watch of changes (increases/decreases) in stock prices on a regular basis, so you can move quickly if you anticipate a change in stock value.

You should also take as little time as possible to make any decision (sell/buy), because the stock market can move quickly, and if you wait too long to make a decision, you may lose money.

We have discussed the possibility of purchasing shares of stock for a corporation you are interested in following .

When purchasing stocks, you must consider all relevant aspects, such as the company’s present financial/economic condition, the country’s financial/economic status, the financial/economic situation of the company’s major consumers, and so on.

This is significant since all of these factors have an impact on stock market pricing.

It is critical to purchase stocks at the correct time for the proper firm, namely when the stock price is low and the stock price is predicted to climb (much) in the near future.

Furthermore, it is critical to sell the stocks at the proper time in order to maximize your profit, specifically when the stock’s value is high and when the stock’s value is projected to decline (significantly) in the near future.

A stock market’s value can fluctuate dramatically over a short period of time, thus it’s critical to maintain watch of changes (increases/decreases) in stock prices on a regular basis, so you can move quickly if you anticipate a change in stock value.

You should also take as little time as possible to make any decision (sell/buy) because the stock market can move quickly, and if you wait too long to make a decision, you may lose money.

Page 57 Problem 3 Answer

It is given that : Survey your classmates and compile a list of questions your class has about stocks. Compile a list of the top five stocks they are interested in.

We have to report our findings.

Answer

When purchasing stocks, you must consider all relevant aspects, such as the company’s present financial/economic condition, the country’s financial/economic status, the financial/economic situation of the company’s major consumers, and so on.

This is significant since all of these factors have an impact on stock market pricing.

It is critical to purchase stocks at the correct time for the proper firm, namely when the stock price is low and the stock price is predicted to climb (much) in the near future.

Furthermore, it is critical to sell the stocks at the proper time in order to maximize your profit, specifically when the stock’s value is high and when the stock’s value is projected to decline (significantly) in the near future.

A stock market’s value can fluctuate dramatically over a short period of time, thus it’s critical to maintain watch of changes (increases/decreases) in stock prices on a regular basis, so you can move quickly if you anticipate a change in stock value.

You should also take as little time as possible to make any decision (sell/buy) because the stock market can move quickly, and if you wait too long to make a decision, you may lose money.

From our findings we get that :

When purchasing stocks, you must consider all relevant aspects, such as the company’s present financial/economic condition, the country’s financial/economic status, the financial/economic situation of the company’s major consumers, and so on.

This is significant since all of these factors have an impact on stock market pricing.

It is critical to purchase stocks at the correct time for the proper firm, namely when the stock price is low and the stock price is predicted to climb (much) in the near future.

Furthermore, it is critical to sell the stocks at the proper time in order to maximize your profit, specifically when the stock’s value is high and when the stock’s value is projected to decline (significantly) in the near future.

A stock market’s value can fluctuate dramatically over a short period of time, thus it’s critical to maintain watch of changes (increases/decreases) in stock prices on a regular basis, so you can move quickly if you anticipate a change in stock value.

You should also take as little time as possible to make any decision (sell/buy), because the stock market can move quickly, and if you wait too long to make a decision, you may lose money.

Page 57 Problem 4 Answer

It is given that : Visit a local bank and ask to speak to one of the representatives about United States Savings Bonds.

Find out about the forms necessary to purchase a bond, the interest it pays, and how long the bonds take to reach their face value.

We have to prepare a report and present our findings to the class.

Answer

When we buy a bond, we’re really buying a debt security since the borrower sells bonds to investors, who then lend money to the borrower for a set length of time.

The borrower will subsequently be responsible for repaying the bond (at an increased value, determined by the interest rate of the bond).

A bond can be purchased through a bank or other financial institution, as well as through the US Treasury’s Treasure Direct website.

You will only need to create an account and put in your personal information if you use the website to buy bonds. All bonds are now available in electronic format.

The current bond interest rate is about0.10%

(in2018). People who bought bonds decades ago, on the other hand, may be receiving interest rates as low as (so bonds are currently less attractive as they were several decades ago).

A bond’s face value can be reached in as little as 20

years (as the treasury promised that you would double your investment in less than20

years, while your initial investment is half of the face value).

By our we findings we get that :

When we buy a bond, we’re really buying a debt security since the borrower sells bonds to investors, who then lend money to the borrower for a set length of time.

The borrower will subsequently be responsible for repaying the bond (at an increased value, determined by the interest rate of the bond).

A bond can be purchased through a bank or other financial institution, as well as through the US Treasury’s Treasure Direct website.

You will only need to create an account and put in your personal information if you use the website to buy bonds.

All bonds are now available in electronic format.

The current bond interest rate is about.10 %(in 2018). People who bought bonds decades ago, on the other hand, may be receiving interest rates as low as (so bonds are currently less attractive as they were several decades ago).

A bond’s face value can be reached in as little as 20 years (as the treasury promised that you would double your investment in less than 20 years, while your initial investment is half of the face value).

Page 58 Problem 5 Answer

The New York Stock Exchange offers many different types of publications.

And, some of these publications are:

NYSE American Options Closing-Only LessOptions fee scheduleOnly \& Batch Extract SpecificationsLate Close ListCMT Fact SheetCMT User request and Authorization FormPenny Pilot ReportTPID List

Furthermore, a monthly report on the impact of Limit Up/Limit Down and Straddle States on options trading is made public each month.

A list of publications that the Exchange offers are :

NYSE American Options Closing-Only LessOptions fee schedule only \& Batch Extract SpecificationsLate Close ListCMT Fact SheetCMT User request and Authorization FormPenny Pilot ReportTPID List

Furthermore, a monthly report on the impact of Limit Up/Limit Down and Straddle States on options trading is made public each month.

Page 58 Problem 6 Answer

Here we have to do, Is to set up of portfolio for a students of 5 to 10 and track the gains and losses of entire month Firstly select students of 5−10

Use stocks like S&P500, Dow Jones We have an investment of 10,000 $

firstly we have to divide the investment for the companies and know how many stocks we own so, we know that in the stock market the profit and loss will vary from the day-day basis

analyze and obtain a value of how much do we get for an entire month and if we didn’t obtain the required repeat until you reach it.

Therefore, Determine the loss and gains as a day-day Comparision and if didn’t obtain any profit then repeat it for an entire month

Page 58 Problem 7 Answer

Given; The payment for a job is to be paid for 30 days of June such that on  2nd day, the payment was 1 cent. 2nd day, the payment was 2 cents. 3rd day, it was 4 cents.

In this way, each day the payment was twice the amount of the previous day.

To find; Draw a grid with six columns and five rows to represent the 30 days

How much do you receive on June 14? June 27? June 30 we have The payment for a job is to be paid for 30  days of June such that on 1st day, the payment was 1 cent.

2nd  day, the payment was 2 cents. 3rd  day, it was 4 cents.

In this way, each day the payment was twice the amount of the previous day.

From the above grid, the payment on June 14 will be 8,192cents.

From the above grid, the payment on June 27 will be 67,108,864 cents.

From the above grid, the payment on June 30 will be 536,870,912 cents.

Hence we cocnluded that From the above grid, the payment on June14 will be 8,192 cents.

From the above grid, the payment on June 27 will be 67,108,864 cents .

From the above grid, the payment on June 30 will be 536,870,912 cents.

Page 59 Problem 8 Answer

Given the money to start the business is 73,000 $ and their ratio is 3:7

To do: Find how much did they invest

The given ratio is  3:7 so let Nick investment is 3x and Matt investment is 7x  then the total investment will be 3x+7x=10x and the total money is 73,000 $

​⇒x=73000/10

=7300

substituting for x we get Nick investment ​3x=3(7300)

=21900​

Matt investment ​7x=7(7300)

=51100​

Therefore, Nick investment is 21900 $

Matt investment is 51100 $

Page 59 Problem 9 Answer

Given the money needed to start a business is 73,000 $ the ratio is 3:7

To do: Percent of the business was owned by Matt

Given the total investment is 3x+7x=10x in this the investment of matt is 7x then the percent will be 7x/10x =0.7 which is equal to 70%

Therefore,Matt owned 70 % of the business

Page 59 Problem 10 Answer

Given that Tom purchased shares of DuPont 47.65 $ per share. He plans to sell the shares when the stock price rises 20%

To do: Find the selling price

Given that there is a 20 % increase to the original price this means that the price is increased by 20% so, the cost is 1.2 then the selling price will be 57.18

Therefore, the selling price is 57.18

Page 59 Problem 11 Answer

Given that The top three shareholders in a certain corporation each own shares of a certain stock.

The corporation’s ownership is represented by a total of x  shares of stock

To do: Express the percent of the corporation owned by the top three shareholders algebraically

Let the shares owned by the three stakeholders be s then the whole stakeholder will be equal to 3s and the share of corporation is x then the percent owned will be

3s(100/x)

=300s/x

Therefore,

the percent owned is 300s/x

Page 59 Problem 12 Answer

Given that Marilyn purchased 2,000  shares of stock for 25.43 $  per share.

She sold them for 44.10 $  per share

To do: Express her capital gain

From the given data we know that total shares 2000 purchased at 25.43 $ Sold at 44.10 $ from this  the total purchase price =50860 the total selling price =88200

Net capital gain = total selling price−total purchase price

=88200−50860

=37340

​and % capital gain is (100∗37340)/50860=73.4%

Therefore, capital gain =73.4%

Page 59 Problem 13 Answer

Given that A local hairdresser bought 450  shares of a cosmetics corporation for 33.50 $  per share. He sold the shares for 39.01 $  per share

To do: From the given data we have number of shares=450

purchase price=33.5

selling price=39.01 then the percentage price​=100(39.01−33.5/33.5)

=16.4 %

​Therefore, the percentage price is 16.4 %

Page 59 Problem 14 Answer

Given that A local hairdresser bought 450  shares of a cosmetics corporation for 33.50 $ per share. He sold the shares for 39.01 $ per share

To do: Find the total purchase price

From the given data we have number of shares=450 purchase price=33.5$

selling price=39.01$ then the total purchase price is ( $33.5)(450) = $15075

Therefore, the total purchase price is  $15075

Page 57 Problem 15 Answer

Given that A local hairdresser bought 450 shares of a cosmetics corporation for 33.50 $ per share. He sold the shares for 39.01 $ per share

To do: find the total selling price

From the given data we have number of shares=450

purchase price=33.5

selling price =39.01

Then total selling price is ​450×39.01

=17,554.50 $

​Therefore, the total selling price is =17,554.50 $

Page 57 Problem 16 Answer

Given that A local hairdresser bought 450 shares of a cosmetics corporation for 33.50 $ per share. He sold the shares for 39.01 $ per share

To do: Find the capital gain

We know that total selling price=17554.5

total purchase price=15075 then the net capital gain will be =17554.5−15075

=2479.5 ​then  % capital gain​=(100∗2479.5)/15075

=16.4 %

​Therefore, the %capital gain16.4%

Page 59 Problem 17 Answer

Given that the Bootle Corporation paid Leslie a quarterly dividend check for 828 $. Leslie owns 450 shares of Bootle

To do: find quarterly dividend for one share of Bootle

From the given data we know that the total number of shares=450 the dividend of all the shares is 828

so, the quarterly dividend for one share of Bootle​=828/450

=1.84

​Therefore, the quarterly dividend for one share of Bootle=1.84

Page 59 Exercise 1 Answer

Given that Aaron owned x  shares of a corporation and received an annual dividend of y dollars

To do: the quarterly dividend for one share

From the given data we have number of shares owned=x annual dividend of shares=ythen the annual dividend of 1 share=y/x then quarterly dividend of 1 share is y/4x

Therefore, the quarterly dividend of 1 share is y/4x

Page 60 Exercise 2 Answer

Given a table consisting of market data

To do: find the difference between the 52−week high and the 52−week low for one share of AT&T

From the given data we have week high of AT &T52

=42.97

week low of AT&T 52=32.95 then the difference is 42.97−32.95=10.02

Therefore, the difference is 10.02

Page 60 Exercise 3 Answer

Given a table consisting of market data

To do: difference between the day’s high and low for one share of Southern Copper

From the table we have Southern copper day high=110.68

Southern copper day low=105.68 then the difference is 110.68−105.68=5

Therefore, the difference between the day’s high and low for one share of Southern Copper=5

Page 60 Exercise 4 Answer

Given a table consisting of market data

To do: find which stock had a close that was furthest from the day’s low

We have the difference between the Southern Copper corp for close and low is  and the difference of is  and difference of is and JPM organ Chase and co.

so, the stocks are IIT Ed Services furthest from day’s low

Therefore, the stocks are IIT Ed Services furthest from day’s low

Page 60 Exercise 5 Answer

Given a table consisting of market data

To do: Determine the close on June 19  for JPM organ Chase

From the given data we get June 19 close= June 20 close−change

June 19 close=37.86+0.79

=38.65

​Therefore, the June19 close=38.65

Page 60 Exercise 6 Answer

Given a table consisting of market data

To do: How many shares of ITT were traded on June 20

From the given data we have shares of ITT trades of 3429.5 then

3429.5(100)=34,29,500 were traded

Therefore, 34,29,500 shares of ITT were traded on June 20

Page 60 Exercise 7 Answer

Given a table consisting of market data

To do: The percent net change from June 19  to June 20  for AT&T

Firstly finding

June 19 close = June 20 close − change

=34.43+0.72

=35.15

then the percentage will be​

=−0.72100/35.15

=−2.048%⇒−2.05%

​Therefore, the percent change is −2.05 %

Page 60 Exercise 8 Answer

Given a table consisting of market data

To do: Find which stock had a day’s high that was approximately 30% less than its 52-week high

 

Given that there is 30 % less than its 52− week high this means70 % of 52 week high this gives 0.7(52) week high and the difference for Southern Copper crop=100.275

for AT&T=30.079

ITT Ed services =92.274 and JP Morgan Chase & Co.=35.336 from this we get

ITT Ed services has 30 % less than it 52−week high

Therefore, ITT Ed services have30 % less than its 52− week high

Page 60 Exercise 9 Answer

Given a table consisting of market data

To do: the difference in the number of shares traded from June 19 to June 20

From the given data we have Shares on June19=59945400

Shares on June20=55377200 and the difference between these two will be 59945400−55377200=4568200

Therefore,the difference between the trades is 4568200

Page 60 Exercise 10 Answer

Given a bar graph of stock’s

To do: What was the day’s open on June 17

From the graph, we could see that the line on June 17 lies at 49.60

so,the opening price will be 49.60 $

Therefore, the opening price on June 17 is 49.60 $

Page 60 Exercise 11 Answer

Given a bar graph of stock

To do: the approximate difference between the day’s high and low on June 18

From the given graph the high June line is at 49.5  and the low line is at 48.5 and the difference between these is 49.5−48.5=1 $

Therefore, the difference between the day’s high and low on June 18 is 1 $

Page 60 Exercise 12 Answer

Given a bar graph of stock’s

To do: On what day was the close also the day’s low

From the given bar graph, we could see that the low edge on a daily basis is marked on June20

So, we can say that the close is low on that day

Therefore, On June20 the close is low

 Page 60 Exercise 13 Answer

Given a bar graph of stock’s

To do: the approximate volume for June 19

From the graph given, on the volume side, we could see that the line on June 19 is marked at 0.75 which means 0.75 millions or 750,000

Therefore, the volume for June 19 is 750,000

Page 61 Exercise 14 Answer

Given that Lea owns 800  shares of ABC  Incorporated. On April 6  the corporation instituted a 5−for−2

stock split. Before the split, each share was worth $42.60

To do: Find shares did Lea hold after the split

Given that the number of shares lea owns=800 after split​=800(5/2)

=2000

​Therefore, 2000  shares Lea hold after the split

Page 61 Exercise 15 Answer

Given that Lea owns 800  shares of ABC, Incorporated. On April 6 the corporation instituted a 5−for−2 stock split. Before the split, each share was worth $42.60

To do: Find the post-split price per share

We know that  Post split price= 2/5 times of pre-split price

⇒2(42.6)/5

=17.04

Therefore, the post-split price is 17.04 per share

Page 61 Exercise 16 Answer

Given that Lea owns 800  shares of ABC, Incorporated. On April 6 the corporation instituted a 5−for−2  stock split.

Before the split, each share was worth $42.60

To do: Show that the split was a monetary non-event for Lea.

From the given data we have number of shares=800

price of share before the split=42.6

total investment before=34080

price share after the split=17.04

Then total investment =34080

So, we can say that split was a monetary non-event for Lea.

Therefore, The split was a monetary non-event for Lea.

Page 61 Exercise 17 Answer

Given a table of closing prices for Microsoft

To do: Find the 3−day moving averages

The 3− day  moving averages will be:

On 23−May 28.05+28.44+28.18/3

=86.67/3

⇒28.22

On 27− May 28.44+28.18+28.31/3

=84.93/3

⇒28.31

On 28− May 28.18+28.31+28.32/3

=84.81/3

⇒28.27

On 29− May 28.31+28.32+27.80/3

=84.43/3

⇒28.14

On 30− May 28.32+27.80+27.31/3

=83.42/3

⇒27.81

On 2− Jun 27.80+27.31+27.54/3

=82.64/3

⇒27.55

On 3− Jun 27.31+27.54+28.30/3

=83.14/3

⇒27.71

On 4− Jun 27.54+28.30+27.49/3

=83.33/3

⇒27.78

On 5− Jun 28.30+27.49+27.71/3

=83.50/3

⇒27.83

On 6− Jun 27.49+27.71+27.89/3

=83.09/3

⇒27.70

On 9− Jun 27.71+27.89+27.12/3

=82.72/3

⇒27.57

On 10− Jun 27.89+27.12+28.24/3

=83.25/3

⇒27.75

On 11− Jun 27.12+28.24+29.07/3

=84.43/3

⇒28.14

On 12− Jun 28.24+29.07+28.93/3

=86.24/3

⇒28.75

Therefore, the 3− day averages are  $28.22, $28.31, $28.27, $28.14, $27.81, $27.55, $27.71, $27.78, $27.83, $27.70, $27.57, $27.75, $28.14, $28.75

Page 61 Exercise 18 Answer

Given a table of closing prices for Microsoft

To do: Find the 10− day moving averages

Finding the averages of 10− days

On 23− May 28.05+28.44+28.18+…+27.49/10

=279.73/10

≈27.97

On 27− May 28.44+28.18+28.31+…+27.71/10

=279.39/10

≈27.94

On 28− May 28.18+28.31+28.32+…+27.89/10

=278.84/10

≈27.88

On 29− May 28.31+28.32+27.80+…+27.12/10

=277.78/10

≈27.78

On 30− May 28.31+28.32+27.80+…+27.12/10

=277.78/10

≈27.78

On 2− Jun 27.80+27.31+27.54+…+29.07/10

=278.46/10

≈27.85

On 3− Jun 27.31+27.54+28.30+…+28.93/10

=279.59/10

≈27.96

Therefore, the averages of 10−day are $27.97,$27.94,$27.88,$27.78,$27.77,$27.85,$27.96

Page 61 Exercise 19 Answer

Given that [email protected]@[email protected], [email protected]

To do: How many shares did Nick buy

From the given data Nick bought 0.67K which means

0.67(1000)

=670 shares

​Therefore, Nick bought 670 shares

Page 61 Exercise 20 Answer

Given that [email protected]@[email protected], [email protected]

To do: +How much did each share cost

From the given data the cost is written after the symbol @ then each share cost rupees 5.01

Therefore, the share cost is 5.01 $

Page 61 Exercise 21 Answer

Given that [email protected]@[email protected], [email protected]

To do: the value of Nick’s trade From the given data we know that number of shares=670 price=5.01  Value of trade = no. of shares ∗ price

substituting the values we get

=(670)(5.01)

=3356.7

​Therefore, the value of trade=3356.7

Page 61 Exercise 22 Answer

Given that [email protected]@[email protected], [email protected]

To do: How many shares did he sell

From the given data,Patrick sold 1.6Kwhich means

1.6(1000)

=1600

​Therefore,

the shares he sold are 1600

Page 61 Exercise 23 Answer

Given that [email protected]@[email protected], [email protected]

To do: For how much did each share sell

The share price is the value written after @ so, each share has sold at 26.14

Therefore, each share was sold at 26.14

Page 61 Exercise 24 Answer

Given that [email protected]@[email protected], [email protected]

To do: Based on Patrick’s sale, what was the closing price of T on the previous trading day

From the given, we know that, The current price is lower than the previous day so Previous day closing =26.14+1.08

=27.22​

Therefore, the closing price is 27.22

Page 61 Exercise 25 Answer

Given that The stock in a real estate corporation was selling for $78  per share with an annual dividend of $1.86

It underwent a 3−for−2 split

To do: What was the value of one share of the stock after the split

From the given data Stock price before split=78

so, the price stock after the split​=2/3(78)

=52​

Therefore, the stock after the split=52

Page 61 Exercise 26 Answer

Given that The stock in a real estate corporation was selling for $78  per share with an annual dividend of $1.86. It underwent a 3−for−2 split

To do: Find the annual dividend after the split

From the given data,it is given that the annual dividend before split=1.86then annual dividend will be after the split​=2/3(1.86)

=1.24​

Therefore,the annual dividend after the split=1.24

Page 61 Exercise 27 Answer

Given;  A stock that was selling for x per share underwent a y−for-p split.

To find: Express the annual dividend after the split algebraically.

Here we have given It was originally paying an annual dividend of d per share annual dividend before split=d it underwent y-for-p split annual dividend after split=p/y∗d=pd/y

Hence the expression for the annual dividend after the split algebraically. is pd/y

Page 61 Exercise 28 Answer

Given; Suki purchased 9,600 worth of stock and paid her broker a 1.75 % broker fee.

She had an immediate need for cash and was forced to sell the stock when it was worth 8,800.

She used a discount broker who charged 32.50  per trade

To find: Compute Suki’s net loss after the broker fees were taken out.

Purchase price: 9,600

Fee purchase: 1.75%

Selling price: 8,800

Fee sale: 32.50

The broker fee is the product of the broker fee rate of 1.75% and the purchase price:

Broker fee purchase =1.75%× Purchase price​

=0.0175×9600

=168

The total purchase cost is the sum of the purchase price and the broker fee of the purchase:

Total purchase cost = Purchase price  Broker fee purchase​

=9600+168

=9768​

The total selling cost is the selling price decreased by the broker fee of the sale (flat rate):

Total selling cost = Selling price − Broker fee sale​

=8800−32.50

=8767.50

The net loss is then the difference between the total purchase cost and the total selling cost:

Net loss = Total purchase cost −Total selling cost​

=9768−8767.50

=1000.50

Thus the net loss is 1000.50.

Hence the net loss is 1000.50. for suki after the broker fees were taken out.

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 119 Problem 1 Answer

Given: Nick friend James Sloan on May 11 for $150.32.

To find: What should he write in the check register and what should the new balance be?

Here is

Hence, We have checked register the new balance be $2349.58.

Page 119 Problem 2 Answer

Given: Nick friend James Sloan on May 11 for $150.32.

To find: We need to determine the balance in his account after star Cable co.

So here is the solution as

Hence, the balance in his account is

Page 120 Problem 3 Answer

Given: Will Rogers quotation is “There have been three great inventions since the beginning of time: fire, the wheel, and central banking”.

To find: What is the Outline in this lesson?

Fire is an essential part of our daily life, because we need to fire to cook food and we also need fire to create warmth (during cold seasons).

The wheel is also an essential part of our daily life, because the wheel made it possible to travel great distances in a short amount of time and vehicles that use wheel have become an essential part of our daily lives.

Similarly,  Central banking is also an essential part of our daily life, because most transactions and payment are done by banks.

So, here is lesson is Central banking is also an essential part of our daily life, because most transactions and payment are done by banks.

Page 120 Problem 4 Answer

Given: Jackie deposited an $865.98 paycheck, a $623 stock dividend check, a $60 rebate check and $130 cash into her checking amount.

Her original balance was $278.91

To find: We need to assume that the checks clear, how many was in her amount after the deposit made ?

This is the starting balance indicated in the problem.

$278.91

278.91+865.98+623+60+130=1,957.89

The values 865.98,623,60 and 130 are added since these amounts which are deposited in Jackie’s account which gives monetary value.

Hence, There was in her amount after the deposit made is $1,957.89

Page 120 Problem 5 Answer       

Given: Rich has t dollars in his checking account. On jane 3, he deposited w, h and v dollars and cashed a check for k dollars.

To find: We need to write an aglebraic expression that represents the amount of money in his account after the transactions.

Lets

Balance account: t dollars

First deposit: w dollars

Second deposit: h dollars

Third deposit: v dollars

Cashed check: k dollars.

Cashing a check implies that you take the amount of the check out of your account and thus your balance of the account will decrease.

The new balance is the old balance of the account increased by all deposits and decreased by the cashed check:

New balance

= Previous balance + Deposits − Cashed check

=t+(w+h+v)−k

=t+w+h+v−k​

Hence,  The aglebraic expression of he amount of money in his account after the transactions is

t+w+h+v−k

Page 120 Problem 6 Answer

Given : John cashed a check for : $ 630

The teller gave him three fifty-dollar bills, eighteen twenty-dollar bills,  t ten-dollar bills.

To find: We need to determine the value of t

So here is

Total amount cashed =$630

No. of 50 dollar bills =3

So, total amount of 50 dollar bills =3×50=$150

Number of 20 dollar bills $=18$

So, total amount of 20 dollar bills =18×20 =$360

No of 10 dollar bills =t

So, total amount of 10 dollar bills =t×10 =10t

Now, 150+360+10t=630

5t=120

i.e. t=12

Hence, We got the value of t is t=12

Page 120 Problem 7 Answer     

Given: Monthly maintenance fee of $13 Check writing fee is $0.07 Last balance check is $289

To find: What was the total of all fees he paid on that account last year ?

Maintenance fee: $13 per month

Check writing fee: $0.07 per check 289 checks We are interest in the total fees in 1 year.

Since there are 12 months in a year, the total maintenance fee is the product of the fee per month and the number of months.

Total maintenance fee = Monthly maintenance fee × Number of months

=13×12 =156

The total check writing fee is the product of the check writing fee per check and the number of checks.

Total check writing fee = Check writing fee per check × Number of checks

​=0.07×289

=20.23

The total fees is then the sum of the total maintenance fee and the total check writing fee:

Total fee = Total maintenance fee + Total check writing fee

=156+20.23

=$176.23

​Hence, In the last year the total of all fees he paid on that account is =$176.23

Page 120 Problem 8 Answer

Given: Joby had $421.56 in her account. after deposit g twenty-dollar and k quarters.

To find: We need to write the expression of the amount of money in her account after the deposit.

So here, Previous balance account: $421.56

Deposits: g20-dollar bills and k quarters

The total value of g20-dollar bills is 20g as each bill is worth 20 dollars.

The total value of k quarters is 0.25k as each quarter is worth 25 cents or 0.25 dollars.

The new balance is the old balance of the account increased by all deposits and the earned interest, and decreased by the written check and the fees that need to be paid:

New balance

= Previous balance + Deposits + Interest − Written checks − Fee

=421.56+(20g+0.25k)

=421.56+20g+0.25k

​Hence, The expression of the amount of money in her account after deposit is

421.56+20g+0.25k

Page 120 Problem 9 Answer

Given: Hector had y dollars in his saving account, Deposit amount twenty-dollar bills and dollar coins.

Total of twenty dollar bills was $60

To find: We need to write the expression for the balance in hector’s account after deposit.

Let’s,Previous balance account: y Deposits: 4 times as many dollar coins as twenty dollar bills Total twenty dollar bills =$60

Let x be the number of twenty dollar bills, then there are 4x dollar coins in the deposit.

The total of the twenty dollar bills is then the product of the number of twenty dollar bills x and the value per bill (20 dollars).

Total twenty dollar bills = Number of twenty dollar × Value per bill

60=x×20

Divide each side by 20/3=x

Thus we know that x=3.

We can then determine the number of dollars coins deposited.

4x=4(3)  =12

Thus the total of the twenty dollars bills was $60,

while the total of the dollar coins is  12$ (as there are 12 dollar coins and each dollar coin is worth  1$ ).

The new balance is the old balance of the account increased by all deposits and the earned interest, and decreased by the written check and the fees that need to be paid:

New balance = Previous balance + Deposits + Interest – Written checks – Fee​

=y+(60+12)

=y+72

​Hence, We got the expression is y+72.

Page 121 Problem 10 Answer

Given: Your balance on 10/29 is $237.47

To find: We need to create a check register for the transactions listed.

So here is

Hence, the register transactions listed as

Page 121 Problem 11 Answer  

Given: Check 115 on 10/29 for $18.00 to Fox high school.

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Problem 12 Answer

Given: Deposit a paycheck for $162.75 on 10/30

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Problem 13 Answer

Given: Your deposit a $25 check for your birthday on 11/4.

To find: We need to create a check register for the transactions listed.

So here

Hence, the register transactions listed

Page 121 Problem 14 Answer

Given: On 11/5, sporting event and run out of money, you use the ATM in the lobby to get $15 for snacks.

To find: Hence, the register transactions listed

So here,

Hence, the register transactions listed

Page 121 Problem 15 Answer

Given: Your credit card bill is due on 11/10, so on 11/7 you write check 116 to Credit USA for $5.16.

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Problem 16 Answer

Given: Your sister repays you $20 on 11/10. you deposit it.

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Problem 17 Answer

Given: You withdraw $25 from the ATM to buy flowers on 11/12.

To find: We need to create a check register for the transactions listed.

So here

Hence, the register transactions listed

Page 121 Problem 18 Answer

Given: You deposit your paycheck for $165.65 on 11/16.

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Problem 19 Answer

Given: Your deposit a late birthday check for $35 on 11/17.

To find: We need to create a check register for the transactions listed.

So here,

Hence, the register transactions listed

Page 121 Exercise 1 Answer

Given:   Ridge wood Saving bank charges a $27

Nancy had $1,400 in her account

July 9, $1,380.15, July 10 $670 and $95.67 July 12

To find: How much will she owe the bank after July 12.

Lets, Overdraft occurs when there are insufficient funds.

Checks are drawn in excess of the remaining funds in the account.

Ddit Balance                                                          ​$1,400.00

July 9 check                                                          $1,380.15

Balance                                                                 $19.85

July 9check                                                           −$745.82

Balance                                                                  (670+95.67)​

July 10 check                                                       $765.67(670+95.67)

Balance                                                                 $765.67

We can see that the July 10 check is overdrawn.

There is no sufficient funds to clear the check.

There is an overdraft.

Thus, the bank will start charging $27 for every check drawn thereafter, including the two July 10 checks.

There are 2 more checks drawn after July 10:

(total of 4 checks)

Overdraft fee =$27×4

=$108

Balance                                                              ​−$745.82

July 11 Check                                                           $130.00

−−−−−−−−−−

Balance                                                                   −$875.82

July 12 check 87

−−−−−−−−−−−−

Ending balance −                                                    $963.42

Overdraft fee                                                       $108.00

Total obligation                                                  −$1,071

Answer: $1,071.42

Hence, The She will owe after 12 July is $1,071.42

Page 121 Exercise 2 Answer

Given: Saving and loan charges a monthly fee of $8.

Overdraft protection fee is $33

Neela’s had balance is $456

Transfer the amount $250

To find: We need to find the statement of her account.

Here is, Overdraft protection fee. =33 dollar

Monthly savings fee. =8 dollar

Total fee charged. =33+8

= 41dollars

actual balance in the account. =256 dollar

Amount deposited. =250 dollar

So, Final balance in the account. =506 dollar

Check amount. =320 dollar

Now, the amount of check and total fees will be deducted from the account

So, final account balance. =506−41−320

=153 dollars

Hence, The account statement show the amount =153 dollars in her account.

Page 122 Exercise 3 Answer

Given: Your Balance on 12/15 is $2,546.50

To find: We need to create a check register for the transactions listed.

So here is the first line of transactions

Hence, The first line of the transactions register is

Page 122 Exercise 4 Answer

Given:  On 12 / 16, you write check 2345  for }$ 54  to Kings Park High School Student Activities.

To find: We need to create a check register for the transactions listed.

So here is the transactions Listed as

Hence, The first line of the transactions register is

Page 122 Exercise 5 Answer

Given:  On 12 / 17,  you deposit your paycheck in the amount of 324.20 .

To find:  We need to create a check register for the transactions listed.

So here the transactions listed as

Hence, The first line of the transactions register is

Page 122 Exercise 6 Answer

Given: Your grandparents send you a holiday check for $ 100 which you deposit into your account on 12 / 20 .

To find: We need to create a check register for the transactions listed.

So here is the transactions as

Hence, The first line of the transactions register

Page 122 Exercise 7 Answer

Given: 2346 to Best Buy in the amount of $326.89, 2347 to Macy’s in the amount of $231.88, and 2348 to Target in the amount of $123.51.

To find: Create a check register for the transactions listed.

Thus, we created the transactions listed.

Page 122 Exercise 8 Answer

Given: As you are writing the check for $301.67, you make a mistake and must void that check. You pay with the next available check in your checkbook.

To find: Create a check register for the transactions listed.

Hence, we created the transactions listed.

Page 122 Exercise 9 Answer

Given: On 12/26, you return a holiday gift. The store gives you $98. You deposit that into your checking account.

To find: Create a check register for the transactions listed.

Hence, we created the transactions listed.

Page 122 Exercise 10 Answer

Given: On 12/28, you write an e-check to Allstate Insurance Company in the amount of $876.00 to pay your car insurance.

To find: Create a check register for the transactions listed.

Hence, we created the transactions listed.

Page 122 Exercise 11 Answer

Given: On 12/29, you withdraw $200 from an ATM. There is a $1.50 charge for using the ATM.

To find: Create a check register for the transactions listed.

 

Hence, we created the transactions listed.

Page 122 Exercise 12 Answer

Given: The data given.

To find: Complete items a through y.

(1) We note that the Item no. for transaction code was 622 in the previous row and 624 in the following row, thus the Item no. for transaction code in this row should be 623 .

(2) We note that the Item no. for transaction code was 628 in the previous row, thus the Item no. for transaction code in this row should be 629 (integer directly following 628 ).

(3) We note that the Item no. for transaction code was 629 in the last row with an Item no. for transaction code, thus the Item no. for transaction code in this row should be 630 (integer directly following 629 ).

(4) We note that the last column of this row contains ” -71|10″, which implies that a payment was made of $ 71.10 and thus we should enter 71 |10 in the cells of (d).

Note: you write -71 at the position of “d.” and 10 in the cell directly to its right. lcolor{default}

(5) We note that this row contains a value in the column “amount of payment or withdrawal” of 500|00, which implies that a payment of $500.00 was made and then we should enter -500|00 in the cells of (e)

(6) This cell should contain the new balance, which is the previous balance of $ 1,792.80 (in previous row/transaction) decreased by the payment of $ 500.00 (part (e)). $ 1,792.80- $ 500.00= $ 1,292.80 Thus we should then enter 1,292|80 into the cells of (f).

(7) We note that this row contains a value in the column “amount of payment or withdrawal” of 51|12, which implies that a payment of $ 51.12 was made and then we should enter -51|12$ in the cells of{g}.

(8) This cell should contain the new balance, which is the previous balance of $ 1,292.80 (result part ()) decreased by the payment of $ 51.12 (part {g}).

$ 1,292.80- $ 51.12= $ 1,241.68

Thus we should then enter 1,241|68 into the cells of {h}.

(9) We note that this row contains a value in the column “amount of payment or withdrawal” of 25|00, which implies that a payment of $ 25.00 was made and then we should enter -25|00 in the cells of (i).

(10) This cell should contain the new balance, which is the previous balance of $ 1,241.68 (result part (11)) decreased by the payment of $ 25.00 (part (i)).

$ 1,241.68- $ 25.00= $ 1,216.68

Thus we should then enter 1,216|68 into the cells of {j}.

(12) We note that this row contains a value in the column “Amount of deposit or interest” of 650|00, which implies that a deposit of $ 650.00 was made and then we should enter +650|00 in the cells of {k}.

(1) This cell should contain the new balance, which is the previous balance of $ 1,216.68 (result part (j)) increased by the deposit of $ 650.00 (part (k)).

$ 1,216.68+ $ 650.00= $ 1,866.68

Thus we should then enter 1,866|68 into the cells of (l).

(13) We note that this row contains a value in the column “amount of payment or withdrawal” of 200|00, which implies that a payment of $ 200.00 was made and then we should enter -200|00 in the cells of {m}.

(14) This cell should contain the new balance, which is the previous balance of $ 1,866.68 (result part (1)) decreased by the payment of $ 200.00 (part {m}).

$ 1,866.68- $ 200.00= $ 1,666.68

Thus we should then enter 1,666|68 into the cells of {n}. Note: you write 1,666 at the position of “n.” and 68 in the cell directly to its right.

(15) We note that this row contains a value in the column “amount of payment or withdrawal” of 90|00, which implies that a payment of $ 90.00 was made and then we should enter -90|00 in the cells of {p}.

(16) This cell should contain the new balance, which is the previous balance of $ 1,666.68 (result part (n)) decreased by the payment of $ 90.00 (part (p)).

$ 1,666.68- $ 90.00= $ 1,576.68

Thus we should then enter 1,576|68 into the cells of {q}.

(17) We note that this row contains a value in the column “amount of payment or withdrawal” of 49|00, which implies that a payment of $ 49.00 was made and then we should enter -49|00 in the cells of {r}.

(18) We note that this row contains a value in the column “amount of payment or withdrawal” of 65|00 which implies that a payment of $ 65.00 was made and then we should enter-65|00 in the cells of (t).

(19) This cell should contain the new balance, which is the previous balance of $ 1,527.68 (result part (20)) decreased by the payment of $ 65.00 (part (t)). $ 1,527.68- $ 65.00= $ 1,462.68 Thus we should then enter 1,462|68 into the cells of (u)

(21) We note that this row contains a value in the column “amount of payment or withdrawal” of 300|00, which implies that a payment of $ 300.00 was made and then we should enter -300|00 in the cells of (v)

(22) This cell should contain the new balance, which is the previous balance of $ 1,462.68 (result part (u)) decreased by the payment of $ 300.00 (part (v)). } $ 1,462.68- $ 300.00= $ 1,162.68 Thus we should then enter 1,162|68 into the cells of (w).

(23) We note that this row contains a value in the column “Amount of deposit or interest” of 400|00, which implies that a deposit of $ 400.00 was made and then we should enter +400|00 in the cells of {x}.

(24) This cell should contain the new balance, which is the previous balance of $ 1,162.68 (result part (w)) increased by the deposit of $ 400.00 (part {x}).

$ 1,162.68+ $ 400.00= $ 1,562.68

Thus we should then enter 1,562|68 into the cells of {y}.

(25) This cell should contain the new balance, which is the previous balance of $ 1,562.68 (result part (y)) decreased by the payment of $ 371.66(3-71|66 is given in the cells directly above the cells of z.).

$ 1,562.68- $ 371.66= $ 1,191.02

Thus we should then enter 1,191|02 into the cells of {z}.

Hence, a complete items is found to be

(1) 623

(2) 629

(3) 630

(4) 71 | 10

(5) -500 | 00

(6) 1,292 | 80

(7) -51 | 12

(8) 1,241 | 68

(9) -25 | 00

(10) 1,216 | 68

(11)+650 | 00

(12) 1,866 | 68

(13) -200 | 00

(14) 1,666 | 68

(15) -90 | 00

(16) 1,576 | 68

(17) -49 | 00

(18)  1,527 | 68

(19) -65 | 00

(20) 1,462| 68

(21) -300| 00

(22) 1,162 | 68

(23)+40

(24) 1,562 | 68

(25) 1,191 | 02

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 25 Problem 1 Answer

Given: The closing prices $28,$31,$37,$38, and $35

To find: 3-day SMA using subtraction and addition method

Solution: Find the average of the first three days Then using that average and using subtraction and addition find the next moving average.

Calculate average closing prices of days 1−3

28+31+37/3 =32

Use subtraction and addition to determine the averages for days 2−4

32−28/3+38/3

=32−9.33+12.67=35.34

Similarly, average for days3−5

35.34−31/3+35/3

=35.34−10.33+11.67/3

day SMA for the closing prices $28,$31,$37,$38,and $35 are $32,$35.34,$36.68

Page 26 Problem 2 Answer

Given: Spreadsheet showing closing prices

To find:

If we add column D to the spreadsheet to calculate the 3-day SMA., then the cell with which we will start and the formula we will use

Solution: We will write the first formula in row D4

We will use the formula of arithmetic mean. The formula in spreadsheet format will be=sum(B2:B4)/3

For the given spreadsheet, to find the 3-day SMA in column D

we will start the formula from the cel lD 4The formula will be=sum(B2:B4)/3

Page 26 Problem 3 Answer

Crossovers signal that a stock trend reversal might be near. Some say that an investor should consider buying when the fast-moving average graph overtakes (rises above) the slow-moving average graph.

Likewise, an investor might consider selling when the fast-moving average graph crosses below the slow-moving average graph.

In the given scenario on the 35th trading day, the 21-day SMA graph rises above the 7-day graph.

This means that the slow-moving average graph overtakes the fast-moving average graph, so the investor should consider selling.

If on the 35th trading day, the 21-day SMA graph rises above the 7-day graph, this indicates that the slow-moving graph has overtaken the fast-moving graph, so the buyer might consider selling the stock.

Page 27 Problem 4 Answer

Given ;

To find :  Why might the author be warning readers to be cautious of averages? How might these words apply to what you have learned?

Here the author want to say that if you borrow money, then this will cause you sorrow.

This means that borrowing money isn’t free and that in order to borrow the money, you will have pay some costs (the interest) in return for the borrowing of the money.

Moreover, in the long run, you will thus have to return the borrowed money plus some additional costs (the interest) and thus in the long run, borrowing money will cost you money.

Hence the author want to convey that in the long run, borrowing money will cost you money.

borrowing money isn’t free

Page 27 Problem 5 Answer

Given: The ten consecutive day closing prices for WalMart Stores Inc are$57.35,$58.61,$57.98,$58.07,$57.50,$56.97,$56.35,$56.83,$57.16,$57.18

To find: The 4-day SMA

Solution: We will find an average of the prices for each of the 4-day time spans: days 1−4, days2−5, days3−6, days4−7, days5−8, days6−9 and days 7−10

We will find the average of the closing prices in groups of four Days

1−4=57.35+58.61+57.98+58.07/4

=232.01/4

=58.00

Days​2−5

​=58.61+57.98+58.07+57.50/4

=232.16/4

=58.04

Days​3−6

​=57.98+58.07+57.50+56.97/4

=230.52/4

=57.63

Days ​4−7

​=58.07+57.50+56.97+56.35/4

=228.89/4

=57.22

Days​ 5−8

​=57.50+56.97+56.35+56.83/4

=227.65/4

=56.91

Days ​6−9

​=56.97+56.35+56.83+57.16/4

=227.31/4

=56.83

Days​7−10

​=56.35+56.83+57.16+57.18/4

=227.52/4

=56.88

​The 4-day SMA for the given ten consecutive day closing prices for WalMart Stores Inc are$58.00,$58.04,$57.63$57.22,$56.91,$56.83,$56.88

Page 27 Problem 6 Answer

Given: The ten consecutive day closing prices for Procter & Gamble Co are$66.21,$65.90,$67.05,$67.03,$66.80,$66.65,$66.65,$65.80,$65.92,$65.21

To find: The 3-day SMA

Solution: We will use the Subtraction and Addition method to find the SMA

Find the average of closing prices in the groups of three

Days1−3=66.21+65.90+67.05/3=66.39

Days2−4=66.39−66.21/3+67.03/3=66.66

Days3−5=66.66−65.90/3+66.80/3 =66.96

Days4−6=66.96−67.05/3+66.65/3=66.83

Days5−7=66.83−67.03/3+66.65/3=66.70

Days6−8=66.70−66.80/3+65.80/3=66.37

Days7−9=66.37−66.65/3+65.92/3=66.12

Days8−10=66.12−66.65/3+65.21/3=65.64

The 3-day SMA for the given ten consecutive day closing prices for Procter & Gamble Co are $66.39,$66.66,$66.96,$66.83,$66.70,$66.37,$66.12,$65.64

Page 27 Problem 7 Answer

Given: The ten consecutive trading day closing prices for International Business Machines Corp are$121.69,$122.85,$120.70,$123.61,$123.18,$122.03,$122.82,$124.14,$124.92,$124.06

To find: The 4-day SMA

Solution: We will use the subtraction and addition method to find SMA

We will find the moving average in the group of four.

Days1−4=121.69+122.85+120.70+123.61/4=122.21

Days2−5=122.21−121.69/4+123.18/4 =122.58

Days3−6=122.58−122.85/4+122.03/4=122.38

Days4−7=122.38−120.70/4+122.82/4=122.91

Days5−8=122.91−123.61/4+124.14/4=123.04

Days6−9=123.04−123.18/4+124.92/4=123.48

Days7−10=123.48−122.03/4+124.06/4=123.99

The 4-day SMA for the given ten consecutive trading day closing prices for International Business Machines Corp are$122.21,$122.58,$122.38,$122.91,$123.04,$123.48,$123.99

Page 27 Problem 8 Answer

Given: The ten consecutive trading day closing prices for Rite Aid Corp are$2.65,$2.63,$2.70,$2.63,$2.50,$2.65,$2.66,$2.56,$2.52,$2.37

To find: The 6-day SMA

Solution: We will use subtraction and addition method to find the SMA

We will find the moving averages in the group of six

Days1−6=2.65+2.63+2.70+2.63+2.50+2.65/6=2.63

Days2−7=2.63−2.65/6+2.66/6=2.63

Days3−8=2.63−2.63/6+2.56/6=2.62

Days4−9=2.62−2.70/6+2.52/6=2.59

Days5−10=2.59−2.63/6+2.37/6=2.54

The 6-day SMA for the ten consecutive trading day closing prices for Rite Aid Corp are$2.63,$2.63,$2.62,$2.59,$2.54

Page 28 Problem 9 Answer

Given: The closing prices for Citigroup Inc are:

To find: The 7-day SMA

Solution: We will rite the data given in a spreadsheet.

Then use the formula for finding SMA from the seventh day.

The 7 day SMA for Citigroup Inc using excel are:

The7 -day SMA for Citigroup Inc for the given data using spreadsheet are:

Page 28 Problem 10 Answer

Given: The closing prices for Dell Inc:

To find: The 10-day SMA

Solution: We will write the data in a spreadsheet Then find SMA by writing formula from tenth day onwards

10-Day SMA for Dell Inc using spreadsheet:

The10 -day SMA for the given data for Dell Inc using a spreadsheet are:

Page 28 Problem 11 Answer

Given: Closing prices

To find: 2-day, 3-day and 5-day

Solution: We will write the data in a spreadsheet then use average formula to find the required moving averages

The 2 day,3 day and 5 day SMA are:

For the given closing prices, the 2 day,3 day and 5 day SMA using a spreadsheet are:

Page 28 Problem 12 Answer

Given: The closing prices for ten consecutive of Yahoo! Inc are:

To find The 2 -day and 5 -day SMA using a spread sheet Graph the closing prices and averages.

The2−day and 5−day SMA are:

The graph showing the closing prices and the average is:

For the given closing prices of Yahoo Inc, The   2−days and 5−days SMA are:

The graph showing the closing price and averages is :

 

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market

Financial Algebra 1st Edition Chapter 2 Modeling a Business

Page 108 Problem 1 Answer

Some states about Touchpad         Company: Hewlett Packard

Year released: 2011

Revenue yr. released: $126.0 billion

Touchpad Introduced in July 2011, the TouchPad was Hewlett Packard′s attempt to compete with Apple′s iPad.

With powerful video capability and impressive Processing speeds, the TouchPad was widely anticipated to be among the only products that could give Apple a run for its money.

Despite large-scale press events and promotions, the HP TouchPad was a colossal failure and was discontinued almost immediately.

As a result of the TouchPad′s failure, the company wrote off $885 million in assets and incurred an additional $755 million in costs to wind down its webOS operations, ending all work on the Touchpad’s failed operating system.

Since then, HP has continued to struggle to maintain its edge in the PC market. The once−dominant PC company is in the midst of a multi−year turnaround plan.

While the plan may have recently begun to bear fruit, investors remain cautious.

Name of the product is : Touchpad

The reason behind its rise and fall are mentioned above. The answers of individuals may vary.

Page 109 Problem 2 Answer

The assessment involves an interview of a local businessperson.

Ask for examples of fixed and variable expenses. Do not ask for amounts as that information is private.

Make a comprehensive list. Also ask about the history of the business and how he or she became involved in it.

The interview of a business person is as follows

The most common fixed expenses are

Rent

Salaries

Electricity bill

Water bill

Natural gas bill

Telephone bill

Transportation

Property expenses

InsuranceInterest expenses etc

Variable Expenses

Raw materials for the production of some product

Credit card fees

Commissions supplies and

production piece rate labor etc

The answer of individuals may vary.

Page 109 Problem 3 Answer

Report is as follows

The top 10  Best Selling Beatles singles are:

She Loves You

Want To Hold Your Hand

Can′t Buy Me Love

I Feel Fine

Day Tripper/We Can Work It Out

Hey Jude

Help!

From Me To You

Hello Goodbye

Get back

According  to some Statistics found online, the Beatles earned approximately 71 million dollars this year and thus it appears that Decca′s decision was a wrong decision as the Beatles were a lot more profitable than expected in 1962.

If a product is released at the wrong time (too early or too late), then the product is more likely to fail.

Decca′s decision was based on the fact that guitar groups were expected to no longer be popular, while this was not the case and thus Decca′s decision was based on the wrong timing.

Answers of individuals may vary.

Page 109 Problem 4 Answer             

Report is as follows

A brand name is a name given to a product or a range of products by the maker of the products.

For this assignment, you will need to get two copies of the same newspaper or magazine.

Then you will need to read through the newspaper/magazine to determine all brand names that are mentioned in the newspaper.

There are many possible brand names that will be mentioned.

For example, some of the most famous brand names are Coca-Cola, Lays, Lego, Burger King, McDonald′s, Google, Yahoo, Amazon, etc.

In general, I suspect that you will find dozens of different brand names in a newspaper or magazine.

Answers of the individuals may vary.

Page 109 Problem 5 Answer

Assessment is as follows

There are many different breakeven calculators available only. If you search for ”Breakeven calculator” using a search engine, then you will find many possible calculators.

For example, some breakeven calculators are:

https://www.calcxml.com/do/breakeven−analysis

https://goodcalculators.com/break−even−calculator/

https://www.omnicalculator.com/business/break−even

The calculator of calc xml requires the total fixed costs, the variable cost per unit, the sales price per unit and the anticipated unit sales.

The calculator will then return how many units you need to sell to cover your costs and your expected profit/loss and the expected number of sold units.

The calculator of good calculators requires the same input. The calculator will calculate your break even units, total costs, net profit and total revenue.

Moreover, the calculator will also create a table of possible units solds and the corresponds costs/profit.

The calculator of omnicalculators require the per unit costs and the fixed costs, but the calculator also allows us to take into account a markup or sell margin.

The calculator will then return us how many units need to be sold to breakeven and returns the corresponding revenue.

The answers of individuals may vary.

Page 109 Problem 6 Answer

Report is as follows

A profit is a financial benefit which is the result of revenue gained from business activity reduced by costs/expenses/taxes.

Profit is calculated as total revenue decreased by total expenses.

However different companies tend to define a profit in different manners.

The difference often lies in what is included in the total expenses/revenue and what is not included.

For example, some companies will include dividends paid out to shareholders as expenses (while other companies won′t), while other companies will not include the interest and taxes in their profit calculations (as they are interested in the profit when not taking into account the interest and taxes), etc.

Answers of individuals may vary.

Page 109 Problem 7 Answer

The fixed costs are costs that are constant no matter how many units you produce. Some possible fixed costs for a lemonade stand are then:

Setting up the stand (such as a table, chairs, etc.).

Creating a sign (such that people know that they can buy lemonade at the stand and how much it costs) etc

The variable costs are costs that depend on how many units you produce. Some possible variable costs for a lemonade stand are then:

Ingredients for lemonade: lemons, water, sugar

Plastic cups (to serve the lemonade in)

Containers for the lemonade (before being served). etc

The price at which you sell your lemonade should cover your fixed and variables costs.

You should determine the  fixed and variables costs per unit (glass of lemonade) by dividing the total costs by expected the number of sold  units (glasses of lemonade).

The price per glass of lemonade should then exceed the total costs per unit (glass of lemonade)

Answers of individuals may vary.

Page 110 Problem 8 Answer

We have given The following bar graph that gives the values of a 1958 Edsel Villager 9-Passenger Station Wagon during the years 1999-2008.

Because the value depends on the condition of the car, car appraisers rate the condition of a classic car.

On this graph, Condition 2 is the best condition and Condition 6 is the poorest condition.

The bar graph is

Here we will examine the bar graph and find the approximate value for the 1958 Villager in a condition of 6 in 2008 .

We note that condition 6 is at the very left side of the bar graph.

And the yellow column shows the condition of the car in 2008.

Thus approximate value for the 1958 Villager in a condition of 6 in 2008  = $800

The answer may vary because be take an approximate value.

The approximate value for the 1958 Villager in a condition of 6 in 2008  = $800

Page 110 Problem 9 Answer

We have given The following bar graph that gives the values of a 1958 Edsel Villager 9-Passenger Station Wagon during the years 1999-2008.

Because the value depends on the condition of the car, car appraisers rate the condition of a classic car.

On this graph, Condition 2 is the best condition and Condition 6 is the poorest condition.

The bar graph is

Here we will examine the bar graph and find the approximate value for the 1958 Villager in a condition of 2 in 2008 .

We note that condition 2 is at the very left side of the bar graph.

And the yellow column shows the condition of the car in 2008.

Thus approximate value for the 1958 Villager in a condition of 2 in 2008  = $14,000

The answer may vary because be take an approximate value.

The approximate value for the 1958 Villager in a condition of 6 in 2008  = $14,000

Page 110 Problem 10 Answer

We have given The following bar graph that gives the values of a 1958 Edsel Villager 9-Passenger Station Wagon during the years 1999-2008.

Because the value depends on the condition of the car, car appraisers rate the condition of a classic car.

On this graph, Condition 2 is the best condition and Condition 6 is the poorest condition.

The bar graph is

Here we will examine the bar graph and find the approximate difference between a 1958 Villager in a condition of 6 and in a condition of 2 in 2008.

From last two part we have

The approximate value for the 1958 Villager in a condition of 6 in 2008  = $800

The approximate value for the 1958 Villager in a condition of 2 in 2008  = $14,000

Thus the approximate difference between a 1958 Villager in a condition of 6 and in a condition of 2 in 2008 is

$14,000− $800

=      $13,200

The answer may vary because be take an approximate value.

The  approximate difference between a 1958 Villager in a condition of 6 and in a condition of 2 in 2008 is  $13,200

Page 110 Problem 11 Answer

We have given The following bar graph that gives the values of a 1958 Edsel Villager 9-Passenger Station Wagon during the years 1999-2008.

Because the value depends on the condition of the car, car appraisers rate the condition of a classic car.

On this graph, Condition 2 is the best condition and Condition 6 is the poorest condition.

The bar graph is

Here we will examine the bar graph and find the approximate difference between a 1958 Villager in a condition of 2  in 1999 and in 2008.

From the bar graph we have

The approximate value for the 1958 Villager in a condition of 2 in 1999 = $7000

The approximate value for the 1958 Villager in a condition of 2 in 2008  = $14,000

Hence difference is = $14,000 −$7000 = $7000

The answer may vary because be take an approximate value.

The  approximate difference between a 1958 Villager in a  condition of 2 in 1999 and in 2008 is  $7000

Page 111 Problem 12 Answer

Given scatter plot is

Here for identifying the correlation we will join the given point and check the slope.

If slope is positive then correlation is “positive correlation”.

If slope is negative then correlation is “negative correlation”.

If the line that going through the data points is horizontal then correlation is “no correlation”.

The scatter plot is

If we sketch a line through the given data points, this line will be increasing (positive slope) and thus the data shows a positive correlation.

The given scatter plot shows “positive correlation”.

Page 111 Problem 13 Answer

We have given scatter plot

Here for identifying the correlation we will join the given point and check the slope.

If slope is positive then correlation is “positive correlation”.

If slope is negative then correlation is “negative correlation”.

If the line that going through the data points is horizontal then correlation is “no correlation”.

The given scatter plot is

If we sketch a line through the given data points, this line will be increasing (negative slope) and thus the data shows negative correlation.

The given scatter plot shows “negative correlation”.

Page 111 Problem 14 Answer

Given scatter plot is

Here for identifying the correlation we will join the given point and check the slope.

If slope is positive then correlation is “positive correlation”.

If slope is negative then correlation is “negative correlation”.

If the line that going through the data points is horizontal then correlation is “no correlation”.

The scatter plot is

If we sketch a line through the given data points, this line will be increasing (positive slope) and thus the data shows a positive correlation.

The given scatter plot shows “no correlation”.

Page 111 Problem 15 Answer

Given scatter plots are:

Need to tell which of the following scatter plots does not show a line of best fit.

As a line of best fit is a single line that best fits the scattered points.

The scatter plot(b) consist two lines segments, thus, consists of scattered points that do not lie on the same line.

Whereas in scatter plot(a) and© all the scattered points that lie on the same line.

For the scatter plots:

The scatter plot(b)  does not show a line of best fit.

Page 111 Problem 16 Answer

Given correlation coefficient,r=0.17 .

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=0.17 is positive, thus, the correlation is positive.

The absolute value of the given correlation coefficient,∣r∣=0.17, is less than 0.3, thus, the correlation is weak.

The correlation coefficient, r=0.17, represents a positive and weak correlation.

Page 111 Problem 17 Aswer

Given correlation coefficient,r=−0.62.

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=−0.62 is negative, thus, the correlation is negative.

The absolute value of the given correlation coefficient is,∣r∣=0.62, and satisfies0.3≤r≤0.75, thus, the correlation is moderate.

The correlation coefficient,r=−0.62, represents a negative and moderate correlation.

Page 111 Problem 18 Answer

Given the correlation coefficient,r=−0.88

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=−0.88 is negative, thus, the correlation is negative.

The absolute value of the given correlation coefficient,∣r∣=0.88 , is more than0.75, thus, the correlation is strong.

The correlation coefficient,r=−0.88 , represents a negative and strong correlation.

Page 111 Problem 19 Answer

Given correlation coefficient,r=0.33

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=0.33 is positive, thus, the correlation is positive.

The absolute value of the given correlation coefficient is,∣r∣=0.33, and satifies 0.3≤r≤0.75, thus, the correlation is moderate.

The correlation coefficient, r=0.33, represents a positive and moderate correlation.

Page 111 Problem 20 Answer

Given correlation coefficient, r=0.49.

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=0.49 is positive, thus, the correlation is positive.

The absolute value of the given correlation coefficient is,∣r∣=0.49 , and satisfies 0.3≤r≤0.75, thus, the correlation is moderate.

The correlation coefficient, r=0.49, represents a positive and moderate correlation.

Page 111 Problem 21 Answer

Given correlation coefficient,r=−0.25.

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=−0.25

is negative, thus, the correlation is negative. The absolute value of the given correlation coefficient,

∣r∣=0.25, is less than 0.3, thus, the correlation is weak.

The correlation coefficient,r=−0.25, represents a negative and weak correlation.

Page 111 Problem 22 Answer

Given correlation coefficient,r=0.91.

Need to describe the given correlation coefficient using the terms strong, moderate, or weak and positive or negative.

As r=0.91 is positive, thus, the correlation is positive.

The absolute value of the given correlation coefficient, ∣r∣=0.91, is more than0.75, thus, the correlation is strong.

The correlation coefficient,r=0.91, represents a positive and strong correlation.

Page 112 Exercise 1 Answer

Given the expense function for a particular item be defined as y=−1,950x+53,000

and the revenue function be defined as y=−450x²+10,000x, where x represents price in each equation.

Graph the two functions using the window:

The graph the two functions using the above window is:

For the expense function for a particular item be defined as y=−1,950x+53,000 and the revenue function be defined as y=−450x²+10,000x.

The graph the two functions using the window:

Page 112 Exercise  2 Answer

Given the expense function for a particular item be defined asy=−1,950x+53,000 and the revenue function be defined as y=−450x²+10,000x, where x represents price in each equation.

Need to determine the x-coordinate and they-coordinate of the first point where the two graphs intersect and round those values to the nearest whole number.

As the two intersect if they have the same value of(x,y). Using this find the required coordinates.

The graph is:

Putting the value of expense finction equals to revenue function gives us:

−1,950x+53,000=−450x²+10,000x

⇒450x²−11,950x+53,000=0​

Using quadratic formula for finding the value of x, gives us:

x=−(−11,950)±√(−11,950)²−4⋅450⋅53,000/2⋅450

=11,950±√142,802,500−95,400,000/900

=11,950±√47,402,500/900

=11,950±6884.94/900

⇒x=11,950+6884.94/900 or

x=11,950−6884.94/900

Solving this further gives us:

x=11,950+6884.94/900

=18,834.64/900

=20.92 and x=11,950−6884.94/900

=5065.06/900

=5.6278≈5.63

The corresponding value of y is y=−1,950x+53,000

=−1,950×(5.6278)+53,000

=42025.79​

The x-coordinate and the y-coordinate of the first point where the two graphs intersect is (5.63,42025.79).

Rounding these values to the nearest whole number gives us(6,42026).

For the expense function for a particular item be defined asy=−1,950x+53,000

and the revenue function be defined asy=−450x²+10,000x, where x

represents price in each equation.

The x-coordinate and the y-coordinate of the first point where the two graphs intersect is (5.63,42025.79).

Rounding these values to the nearest whole number gives us(6,42026)

for the graph:

Page 112 Exercise  3 Answer

Given the expense function for a particular item be defined as y=−1,950x+53,000

and the revenue function be defined as y=−450x²+10,000x, where xrepresents price in each equation.

Need to tell the significance of the point where the two graphs intersect in the context of expense, revenue, and price.

As at the point of intersection the expense function is equal to the revenue function or the price at which the profit is zero.

Thus, the point is called the break even point for the graph:

For the expense function for a particular item be defined as y=−1,950x+53,000 and the revenue function be defined as y=−450x²+10,000x, where x represents price in each equation.

The point where the two graphs intersect in the context of expense, revenue, and price is the breakeven point for the graph:

Page 112 Exercise  4 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To determine what will happen to the quantity demanded as the price of the eyePOD increases.

We need to use the fact that the demand for a product depends on the quality of the product as well as the price of the product.

We know that the demand for a product depends on the quality of the product as well as the price of the product.

Thus, in the case of eyePOD, if the price of the eyePOD increases then certainly the quantity of the product demanded will decrease.

Hence, as the price of the eyePOD increases, the demand for the EyePOD will decrease.

Page 113 Exercise  5 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To make a scatterplot of the points represented by(p,q) which are given by​

(300,10,000),(325,8,900),(350,8,800),(375,8,650),(400,6,700)

(425,6,500),(450,5,000),(475,4,500),(500,4,450),(525,3,000)

We need to plot the above-given coordinates on the x,y graph to get the scatterplot.

On plotting prices on x−axis  and quantity on y−axis, we get:

Hence, the scatter plot of the data is given by

Page 113 Exercise  6 Answer

Given, a company is interested in producing and selling a new device called an eye POD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To determine the correlation coefficient and to determine whether this line is a good predictor or not.

We need to calculate the correlation coefficient through LinReg(ax+b)L1,L2.

On calculating correlation coefficient through Lin Reg(ax+b)L1,L2, we get:

y=ax+b

a=−30.7393939394

b=19330

r2=0.9659385288

r=−0.9828217177​

Therefore, the correlation coefficient is −0.98 which represents a negative correlation.

As the correlation is close to ±1, there is a strong correlation.

Thus, we conclude that the line is a good predictor since there is a strong, negative correlation between the price and the demand.

Hence, the correlation coefficient is −0.98 and yes, the regression line is a good .

Page 113 Exercise  7 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To write the regression equation. We need to use Lin Reg(ax+b)L1, L2 to get the equation and then replace x,y with p,q respectively.

Using Lin Reg(ax+b)L1,L2 to get the regression equation,

y​=ax+b

a=−30.7393939394

b=19330

r2=0.9659385288

r=−0.9828217177​

Thus, the regression equation is given by y=ax+b=−30.74x+19330​

On replacing x,y with p,q respectively, we get: q=−30.74p+19330 which is represented by

Hence, the regression equation is q=−30.74p+19330 which is represented by

Page 113 Exercise  8 Answer

Given, The accounting department has calculated that this could be the biggest product to hit the market in years.

It anticipates the fixed costs to be $160,000 and the variable cost to be $150 per eye POD.

To  Express expenses E,  as a function of q, the quantity produced.

We need to form an equation with two variables using the given statement.

We have, the fixed cost=$160,000 the variable cost=$150 per eye POD let q represent the number of eye PODs, then the expenses E  can be expressed as

E=150q+160,000

Hence, expenses E,  as a function of q , the quantity produced can be expressed as

E=150q+160,000

Page 113 Exercise  9 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

The horizontal axis represents price, and the vertical axis represents quantity.To express the revenue R, in terms of p and q.

We need to use the fact that the product of the price and the quantity gives the revenue of the product.

We know that the product of the price p and the quantity q gives the revenue R.

Thus, we conclude that the revenue R can be expressed as R=pq

Hence, the revenue R, in terms of p and q can be expressed as R=pq

Page 113 Exercise  10 Answer

Given, A company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To express the revenue R,  in terms of p.

We need to use the fact that the product of the price and the quantity gives the revenue of the product.

We know that the product of the price p and the quantity q gives the revenue R.

Thus, we conclude that the revenue R can be expressed as R=pq

On replacing q with−30.74p+19,330, we get:

R=p(−30.74p+19,330)=−30.74p{2}+19,330p

R=−30.74p{2}+19,330p

​Hence, the revenue R, in terms of p can be expressed as R=−30.74p{2}+19,330p

Page 113 Exercise 11 Answer

Given, A company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To express expenses E,  in terms of p.

We need to use the transitive property of dependence and make an equation of two variables.

From the previous exercise15 , the expenses E, in terms of q can be expressed as

E=150q+160,000

Now, on replacing q with −30.74p+19,330, we get:

E=150(−30.74p+19,330)+160,000

=−4,611p+2,899,500+160,000

=−4,611p+3,059,500

​Hence, the expenses E,  in terms of p can be expressed as

E=−4,611p+3,059,500

Page 113 Exercise 12 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To determine an appropriate maximum horizontal-axis value.

We need to put zero for the expense function to get the maximum horizontal-axis value.

We have, expense function E=−4,611p+3,059,500 and revenue function R=−30.74p{2}+19,330p from the previous exercise.

On putting expense function as zero, we get:

0=−4,611p+3,059,500

4,611p=3,059,500

p=663.52≈700​

Hence, an appropriate maximum horizontal-axis value is 700.

Page 113 Exercise 13 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To determine an appropriate maximum vertical-axis value.

We need to put zero for the price at the expense function to get the maximum vertical-axis value.

We have, expense function E=−4,611p+3,059,500 and

revenue function R=−30.74p{2}+19,330p from the previous exercise.

On putting zero for the price at the expense function, we get:

E=3,059,500≈3,100,000​

Hence, an appropriate maximum vertical-axis value is 3,100,000.

Page 113 Exercise 14 Answer

Given, A company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To draw a graph of the expense and revenue functions.

We need to get the (x,y) coordinates of the function to draw the graph.

On drawing expense function, represented by a black line and revenue function, represented by purple parabola, we get:

Hence, the graph of the expense and revenue functions is given by

Page 113 Exercise  15 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To determine the coordinates of the maximum point on the revenue graph.

We need to use the fact the maximum of a quadratic function y=ax{2}+bx+c  lies at the symmetry x=−b/2a.

We have the revenue function from the previous exercise R=−30.74p{2}+19,330p.

Since we know that the maximum of a quadratic function y=ax{2}+bx+c lies at symmetry

x=−b/2a.

So, here is a=−30.74

b=19,330 and c=0​

Now, ​p=−b/2a

=−19330/2(−30.74)

≈314.41

​Therefore, the maximum revenue price is $314.41

On putting this value of p in revenue function, we get: R=3,038,784.16 which represents the maximum revenue.

Hence, the coordinates of the maximum point on the revenue graph is

and the graph of the revenue function is given by

Page 113 Exercise  16 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To express the profit, P in terms of p.

We need to find the difference between the revenue and the expense function to get the profit function.

We have the revenue and the expense function from the previous exercises as

​E=−4,611p+3,059,500

R=−30.74p²+19,330p

Since the difference between the revenue and the expense function gives the profit function, we get:

P=R−E

=−30.74p{2}+23,941p−3,059,500

​Hence, the profit function P in terms of price p is represented as

=−30.74p{2}+23,941p−3,059,500

Page 113 Exercise  17 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eye POD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To plot the profit function and to determine the coordinates of the maximum point of the profit graph and determine when the price p is profit maximized.

We need to use the fact the maximum of a quadratic functiony=ax{2}+bx+c lies at the symmetry x=−b/2a.

We have the profit function from the previous exercise P=−30.74p{2}+23,941p−3,059,500

Since we know that the maximum of a quadratic functiony=ax{2}+bx+c lies at symmetry

x=−b/2a

So, here is a=−30.74

b=23,941 and

c=−3,059,500

​Now, ​p=−b/2a

=−23,941/2(−30.74)

≈314.41

​Therefore, the maximum revenue price is$314.41

Now, from the graph, we conclude that maximum profit is made at a price $389.41

Also, the coordinates of the maximum point of the profit graph is

Hence, the coorinates of the maximum point of he profit graph is and the profit is maximized at $389.41.

Page 113 Exercise  18 Answer

Given, a company is interested in producing and selling a new device called an eyePOD (eyewear personal optical device).

The eyePOD is an MP3 and video player built into a pair of sunglasses.

The user can listen to music from the small earphones and watch videos projected on the screen behind the glasses.

To find the shares that must be sold to get enough money to start the business if shares of stock are sold with an initial value of $10.

We need to use the maximum profit price for business profits.

We have the expense function from the previous exerciseE=−4,611p+3,059,500

Also, the maximum profit price is  $389.41

Thus, the expense is given by

​E=−4,611(389.41)+3,059,500

≈$1,263,930.49 which represents the total income of the shares.

Since the cost of each shares is$10. Consider x shares then the total income of the shares is 10x which is given by

10x=1,263,930.49

x=126,393.049

Hence, the number of shares that must be sold to get enough money to start the business is 126,393.049.

 

Chapter 2 Solving Linear Inequalities

• • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.1 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.2 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.3 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.4 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.5 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.6 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.7 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.8 Modeling a Business