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Go Math! The Practice of Statistics for the AP Exam 5th Edition Chapter 7 Sampling Distributions Solutions Introduction Answer Key

Introduction Chapter 7: Sampling Distributions Solutions 

In this chapter, students will find out about What Is a Sampling Distribution, Sample Proportions, and Sample Mean. It also consists of a Chapter review, exercise review, and records exercise test. By the end of the section, students will have the ability to:

1. First, distinguish between a parameter and a statistic.

2. Use the sampling distribution of a statistic to assess a declare approximately a parameter.

3. Distinguish the distribution of a populace, issuance of a pattern, and the sampling distribution of a statistic.

4. Specify whether or not now no longer a statistic is an independent estimator of a populace parameter.

5. Describe the connection between pattern length and the variety of a statistic.

6. Find the suggested and trendy deviation of the sampling distribution of a pattern percentage p^. Check the 10% circumstance earlier than calculating sp^.

7. Determine if the sampling distribution of p^ is about Normal.

8. If appropriate, use a Normal distribution to calculate chances concerning p^.

9. Find the suggested and trendy deviation of the sampling distribution of a pattern meant x. Check the 10% circumstance earlier than calculating sigma x

10. Explain how the sampling distribution of x is suffering from the form of the populace distribution and the pattern length.

11. If appropriate, use a Normal distribution to calculate chances concerning x.

The sampling distribution is a statistical process of interpreting the data produced by random sampling. As the random samples have an element of chances, the laws of probability are used to certify the amount of the chances conclusions drawn from sampling distributions will hold valid.

In this chapter, students shall learn the following:

The basic concept of sampling distributions

Description of sampling distributions

Sampling distribution of sample proportion

Sampling distribution of the sample means

Central Limit theorem for the sampling distribution of sample mean

Learn about the basic concept of sampling distributions.

Get to know the description of sampling distributions.

Understand parameters of statistics.

The sample variance is a measure of the spread of a sampling distribution.

The shape of a sampling distribution may be left-skewed, right-skewed, roughly symmetric, and even approximately normal.

This exercises section will teach will students about sampling distribution. They will begin the learning journey by describing sampling distributions. To trust a statistic for estimating a parameter, its sampling distribution must be analyzed for shape, center, and spread. Then, they will understand what the center of the sampling distribution is. If the mean of a sampling distribution is equal to the parameter’s estimated value, then such statistics is an unbiased estimator. Otherwise, it is a biased estimator. Lastly, they will observe the spread of the sampling distribution in this section.

SLearn to solve problems based on the sample proportion.

Understand sampling distribution of sample proportion.

Perform activities related to the topics above.

The exercises section will focus on understanding the sample proportion of successes that are closely related to x The sampling distribution and mean of sample proportion measure the proportion of success. The sampling distribution of proportion follows the binomial probability law if the random sample of ‘n’ is obtained with replacement.

Learn about the sampling distribution of the sample mean.

Understand the center and spread of the sample mean.

Solve problems based on the sample mean.

In the section, students will begin learning about the sample mean, which is an average of a data set. The sample mean can be used to calculate the central tendency, standard deviation, and variance. Furthermore, students can apply the sample mean to many uses, including calculating population averages. The following results are obtained from the exercises of the sampling distribution of the sample mean.

Lastly, they will understand that the sample mean has a sampling distribution that is (approximately) normal with a mean equal to the population mean for X and a standard deviation equal to a standard deviation of X divided by the square root of the sample size.

Implement the concept of sampling distribution using a case study.

Solve problems based on the case study.

Understand the concept of the sampling distribution.

Details of the case study

The chapter-opening Case Study described a sampling distribution problem in “Building Better Batteries”. Here, a randomly selected battery sample is drained to find the hours till it works and the data is provided. The data mean and standard deviation will give the center and spread of the distribution. It is an unbiased estimator if the sample mean is near the population (17 givens). If the standard deviation value is small, its spread is small and will follow a normal distribution.

Implement the concept of sampling distribution using a case study.

Gain insight into case study-based questions.

Solve questions based on acquired knowledge.

The sample variance is a measure of the spread of a sampling distribution.

The shape of a sampling distribution may be left-skewed, right-skewed, roughly symmetric, and even approximately normal.

 

Learn about the sampling distribution of sample proportion.

Study the shape, center, and spread of sample proportion.

Solve problems on the above topics.

The exercises of the sampling distribution of sample proportion are studied like this. The following results can be observed from the sampling distribution p^ by shape, which is roughly symmetric and somewhat bell-shaped (Approximate normal distribution). It seems to depend on n and, the center, the mean of p^ is quite close to p (Unbiased estimator), and the spread, standard deviation of p^ is very small. It depends on n and p.

Learn to solve problems based on the sample proportion.

Understand sampling distribution of sample proportion.

Perform activities related to the topics above.

The exercises section will focus on understanding the sample proportion of successes that are closely related to x The sampling distribution and mean of sample proportion measure the proportion of success. The sampling distribution of proportion follows the binomial probability law if the random sample of ‘n’ is obtained with replacement.

Learn about the sampling distribution of the sample mean.

Understand the center and spread of the sample mean.

Solve problems based on the sample mean.

In the section, students will begin learning about the sample mean, which is an average of a data set. The sample mean can be used to calculate the central tendency, standard deviation, and variance. Furthermore, students can apply the sample mean to many uses, including calculating population averages. The following results are obtained from the exercises of the sampling distribution of the sample mean.

Lastly, they will understand that the sample mean has a sampling distribution that is (approximately) normal with a mean equal to the population mean for X and a standard deviation equal to a standard deviation of X divided by the square root of the sample size.

Implement the concept of sampling distribution using a case study.

Solve problems based on the case study.

Understand the concept of the sampling distribution.

Details of the case study
The chapter-opening Case Study described a sampling distribution problem in “Building Better Batteries”. Here, a randomly selected battery sample is drained to find the hours till it works and the data is provided. The data mean and standard deviation will give the center and spread of the distribution. It is an unbiased estimator if the sample mean is near the population (17 givens). If the standard deviation value is small, its spread is small and will follow a normal distribution.

Implement the concept of sampling distribution using a case study.

Gain insight into case study-based questions.

Solve questions based on acquired knowledge.

In the exercises section, students will solve problems based on sampling distributions with the help of reliable case studies and understand the background of each problem encountered within the said case study. The exercises are centered around the distribution samples, sample means, and the formula needed to solve them efficiently. It is further explained through the existing knowledge about distributions and samples from earlier chapters.

Learn to implement the concept of sampling distribution in a situation.

Solve problems using the spread, center, and shape of the distribution.

Understand random sample.

Implementation of experimental study
This Frappy section is modeled after actual AP ® Statistics exam free-response questions. It contains inquiries related to the sampling distribution. In the question, a population of 2500 students is given for which population mean, and population standard deviation are known. Then, a random sample of 50 students is selected for which we have to draw the sampling distribution. It can be done after determining the mean and standard deviation of the sample, which will provide information about the center and spread, respectively. After which, everything can be solved with the method discussed above.

Practice on some detailed problems to brush the concepts you have learned through the journey of this chapter.

Solve problems on key topics of the lesson.

Under in-depth the concept of population and sample.

Some of the brief concepts studied under Review Exercises are where the parameter defines the character of the population, and statistics determines the characteristics of the sample. Students must solve problems on how to draw a sample distribution if its center, spread, and shape are known, which is dealt with while analyzing a problem. They will also revise how sampling distribution of mean and sample proportion can be drawn similarly.

Practice on some objective and numerical problems to brush the concepts you have learned through the journey of this chapter.

Understand the sampling distribution of the sample mean.

Solve problems on the above concept.

This extra AP® Statistics Practice Test exercises section contains multiple-choice questions and numerical questions associated with the basic concept of sampling distributions, the description of sampling distributions, the sampling distribution of sample proportion, the sampling distribution of the sample mean, and the Central Limit theorem for the sampling distribution of the sample mean that are already discussed above

Practice on some objective and detailed problems to brush the concepts you have learned through the journey of three chapters.Solve problems on discrete and continuous random variables. Understand the Central Limit theorem for the sampling distribution of the sample mean.

The Cumulative AP® Practice Test 2 section contains multiple-choice questions and numerical questions associated with randomness, probability, and simulation, probability rules, conditional probability, and independence, discrete and continuous random variables, transforming and combining random variables, binomial and geometric random variables, the basic concept of sampling distributions, description of sampling distributions, sampling distribution of sample proportion, sampling distribution of the sample mean, and central limit theorem for the sampling distribution of the sample mean that are already discussed in the chapters altogether.

Carnegie Learning Algebra I Student Text Volume 1 3rd Edition Chapter 4 Sequences

Page 224 Problem 1 Answer

We have been given to Carefully cut out sequences A through P. We have been asked to make sure you do not cut away the letter representing the sequence.By using the snipping tool, we will cut the sequences A through P and paste here one-by-one.

Sequence A-

Sequence B-

Sequence C-

Sequence D-

Sequence E-

Sequence F-

Sequence G-

Sequence H-

Sequence I-

Sequence J-

Sequence K-

Sequence L-

Sequence M-

Sequence N-

Sequence O-

Sequence P-

The cutout of the Sequences A through P by ensuring that no sequence cut away the letter is-

Sequence A-

Sequence B-

Sequence C-

Sequence D-

Sequence E-

Sequence F-

Sequence G-

Sequence H-

Sequence I-

Sequence J-

Sequence K-

Sequence L-

Sequence M-

Sequence N-

Sequence O-

Sequence P-

Page 224 Problem 2 Answer

We have been given to determine the unknown terms of each sequence.We have been asked to describe the pattern under each sequence.We will use the proprieties of sequence.

45,90,180,360,…

The solution goes this way:

45,45×2=90,92×2=180,180×2=360

The sequence follow the pattern of multiplication of 2 to the previous term to get to the next successive term.

So, a5=360×2=720

a6=720×2=1440

a7=1440×2=2880

The next numbers in the sequence are 720,1440,2880.

−4,−2,0,−2,…

In this sequence, the terms are proceeding further in +2 pattern. And thus following Arithmatic Progression.

a5=a4+2=2+2=4

a6=a5+2=4+2=6

a7=a6+2=6+2=8

The next numbers in the sequence are 4,6,8−2,−6,−18,−54,…

The solution goes this way:−2,−2×3=−6,−6×3=−18,−18×3=−54…and so on.

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by multiplying the previous term with 3 and so on respectively.

Thus, the next numbers should be in this pattern:

a5=−54×3=−162

a6=−162×3=−486

a7=−486×3=−1458

​The next numbers in the given sequence are −162,−486,−1458

2,5,10,17…and so on

The solution goes this way: 2,2+3=5,5+5=10,10+7=17,… and so on.

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by adding successive numbers 3,5,7,9,11… and so on respectively, which are odd numbers.

Thus, the next numbers should in this pattern:

a5=17+9=26

a6=26+11=37

a7=37+13=50

The next numbers in the sequences are 26,37,50 E.4,7/4,1/2,−11/4.…

The solution goes this way:4,4−9/4=7/4,7/4−9/4=−2/4

=−1/2,−1/2−9/4

=−11/4 and so on.

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by subtracting −9/4 from the previous term.

Thus, the next numbers should be in this pattern

a5=−11/4−9/4

=−20/4

a6=−20/4−9/4

=−29/4

a7=−29/4−9/4

=−38/4

The next numbers in the sequence are −20/4,−29/4,−38/4.

123. 1234,123.4,12.34,1.234,…

The solution goes this way:

1234,1234/10

=123.4,123.4/10

=12.34,12.34/10

=1.234… and so on

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by dividing the previous term by 1/10.

Thus, the next numbers should be in this pattern:

a5=1.234/10

=0.1234

a6=0.1234/10

=0.01234/a7

=0.01234/10

=0.001234

The next numbers are 0.1234,0.01234,0.001234

1,−2,3,−4,5,…

Firstly, we note that the absolute value of the terms increase by 1.

1,2,3,4,5,6,7,8,9,…

So, we can show the absolute value fo the nth term by n.

Secondly, we observed that the signs (+,−) of each term alternates, and we start with a positive term:+,−,+,−,+,−,+,−,+,−,+,−….

And, we can achieve the correct sign for the nth term by (−1)n+1

If we follow the above conclusions we get that,

The sign for fifth term is + and the term for it is 4+1=5

The sign for sixth term is −and the term for it is−(5+1)=−6

The sign for seventh term is + and the term for it is6+1=7

The next term of the sequence are 5,−6,7−20,−16,−12,−8,−4,…

In this sequence, the terms are proceeding further in +4 pattern.

Thus,

a5=a4+4=−4+4=0

a6=a5+4=0+4=4

a7=a6+4=4+4=8​

The next terms are 0,4,8

1,10,100,1000,…

The solution goes this way: 1,1×10=10,10×10=100,100×10=1000 and so on

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by multiplying 10 respectively

Thus, the next numbers should be in this pattern :

a5=1000×10=10000

a6=10000×10=100000

a7=100000×10=1000000

Thus, the next numbers are 10000,100000,1000000.

−5,−5/2,−5/4,−5/8.…

The solution goes this way:

−5,−5×1/2=−5/2,−5/2×1/2

=−5/4,−5/4×1/2

=−5/8….. and so on

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by multiplying by 1/2 respectively

Thus, the next numbers should be in this pattern :

a5=−5/8×1/2=−5/16

a6=−5/16×1/2=−5/32

a7=−5/32×1/2=−5/64

The next terms in the sequence are −5/16,−5/32,−5/64,5,5,3.5,2,…

The solution is as follows: 6−5−1.5=5,5−1.5=3.5,3.5−1.5=2…and so on.

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by subtracting  1.5 respectively from the previous term

Thus, the next numbers should be in this pattern :

a5=2−1.5=0.5

a6=0.5−1.5=−1

a7=−1−1.5=−2.5

The next terms in the given sequence are0.5,−1,−2.5

86,85,83,80,76,…

The solution is as followed: 86,86−1=85,85−2=83,83−3=80,80−4=76… and so on

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by subtracting successive numbers 1,2,3,4… and so on respectively, which are natural numbers.

Thus, the next numbers shoule be in this pattern

a5=76−5=71

a6=71−6=65

a7=65−7=58​

The next numbers in the sequence are 71,65,58

M.−16,4,−1,1/4.…

The pattern goes this way:−16,−16×(−1/4)=4,4×(−1/4)=−1,−1×(−1/4)=1/4 and so on

Firstly we note that the numbers are multiplied by−1/4 respectively from each previuos term.

Thus, the next numbers should be in this pattern

a5=1/4×(−1/4)=−1/16

a6=−1/16×(−1/4)=1/64

a7=1/64×(−1/4)=−1/256

The next numbers in the sequence are −1/16,1/64−1/256.

1473. 1473.2,1452.7,1432.2,1411.7,….

The solution is as followed:14732.2−2.5=1452.7,14527−20.5=1432.2,1432.2−20.5=1411.7 and so on

If we analyse this we find that the numbers in the sequence follows a certain pattern that is generated by subtracting 20.5 from each previous number.

Thus, the next numbers should be in this pattern:

a5=1411.7−20.5=1391.2

a6=1391.2−20.5=1370.7

a7=1370.7−20.5=1350.2​

The next terms in the given sequence are 1391.2,1370.7,1350.2.

O.√5,2,√3,√2,1,0,√−1.…

The pattern goes this way: √5,√4=2,√3,√2,√1=1 and so on…

We observed that the pattern is the square root of natural numbers in receding order from 5.

Thus, the next numbers should be in this pattern :

a5=√−2

a6=√−3

a7=√−4

The next numbers in the sequence are  √−2,√−3,√−4−4,12,−36,108,….

The solutions is as followed:−4,−4×(−3)=12,12×(−3)=−36,−36×(−3)=108 and so on Firstly, we observed that the previous numbers are multipled by -3 to get the next successive term.

Thus, the next numbers should be in this pattern:

a5=108×(−3)=−324

a6=−324×(−3)=972

a7=972×(−3)=−2916

The next terms in the given sequence are −324,972,−2916.

The next three terms for all the above given sequences are as follows:

A.720,1440,2880

B.4,6,8

C.−162,−486,−1458

D.26,37,50

E.−20/4,−29/4,−38/4.

F.0.1234,0.01234,0.001234

G.5,−6,7

H.0,4,8

I.10000,100000,1000000.

J.−5/16,−5/32,−5/64

K.0.5,−1,−2.5

L.71,65,58

M.−1/16,1/64,−1/256.

N.1391.2,1370.7,1350.2.

O.√−2,√−3,√−4

P.−324,972,−2916.

Page 224 Problem 3 Answer

We have been given that sort the sequences into groups based on common characteristics.

We have been asked to list the letters of the sequences in each group and To provide a rationale as to why you created each group.

We will  be Using the properties of sequence.

A sequence is an ordered list of objects. Like a set, it contains members (also called elements or terms).

The number of ordered elements (possibly infinite) is called the length of the sequence.

Unlike a set, order matters, and a particular term can appear multiple times at different positions in the sequence.

The sequence are grouped as

Adding/Subtracting: B,E,H,K,N

Multiplying/Dividing: A,C,F,I,J,M,P

Other Pattern: D,G,L,O

The sequence is an ordered list of elements, possibly infinite in length.

The sequence is grouped as:

Adding/Subtracting: B,E,H,K,N

Multiplying/Dividing: A,C,F,I,J,M,P

Other Pattern: D,G,L,O

Page 224 Problem 4 Answer

We are given a sequence 45,90,180,360,​_,____,____​.

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence45,90,180,360,____,____

The common ratio that we can observe here is 90/45

=2/180

90=2,so on

​From the given sequence we can observe that the value of a second term can be obtained by multiplying 2 with the first term.

Similarly, the other terms can be obtained by multiplying 2 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 45,90,180,360,_,__,__is multiplication

Page 224 Problem 5 Answer

We are given a sequence−4,−2,0,2,___,___We have to tell which mathematical operation can be performed in order to find the next terms of sequence

We will be using the mathematical operation concept to tell the required operation

We are given a sequence−4,−2,0,2,___,___

The common difference that we can observe here is −2−(−4)=2

0−(−2)=2,so on

From the given sequence we can observe that the value of a second term can be obtained by adding 2 with the first term.

Similarly, the other terms can be obtained by adding 2 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence −4,−2,0,2,____,_____,___is addition

Page 224 Problem 6 Answer

We are given a sequence−2,−6,−18,−54,______,______We have to tell which mathematical operation can be performed in order to find the next terms of sequence We will be using the mathematical operation concept to tell the required operation

We are given a sequence−2,−6,−18,−54,______,______

The common ratio that we can observe here is −6/−2=3/−18−6=3,​so​on

​From the given sequence we can observe that the value of a second term can be obtained by multiplying 3 with the first term.

Similarly, the other terms can be obtained by multiplying 3 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence−2,−6,−18,−54,___,____,____ is multiplication

Page 224 Problem 7 Answer

We are given a sequence2,5,10,17,__,__,____,_____

We have to tell which mathematical operation can be performed in order to find the next terms of sequence We will be using the mathematical operation concept to tell the required operation

We are given a sequence 2,5,10,17,____,_____,_____

We can observe here the difference between the numbers is odd numbers i.e.

5−2=3

10−5=5

17−10=7,so on​

From the given sequence we can observe that the value of a second term can be obtained by adding an odd number starting with 3 to the first term.

Similarly, the other terms can be obtained by adding odd numbers with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 2,5,10,17,____,_____,_____ is addition 

Page 224 Problem 8 Answer

We are given a sequence 4,7/4,−1/2,−11/4

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

From the given sequence we can observe that the value of a second term can be obtained by subtracting 9/4 with the first term.

Similarly, the other terms can be obtained by subtracting 9/4 with the previous term.

We are given a sequence 4,7/4,−1/2,−11/4

We can observe the common difference as

4−7/4=9/4

7/4−(−1/2)=9/4,so on

From the given sequence we can observe that the value of a second term can be obtained by subtracting with the first term.

Similarly, the other terms can be obtained by subtracting with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 4,7/4,−1/2,−11/4 is the subtraction.

Page 224 Problem 9 Answer

We are given a sequence1234,123.4,12.34,1.234,____,___

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation

We are given a sequence1234,123.4,12.34,1.234,_____,______

Here, we can observe a common ratio of 123.4/1234=1/10

12.34/123.4=1/10,so on

​From the given sequence we can observe that the value of a second term can be obtained by multiplying with 1/10 to the first term.

Similarly, the other terms can be obtained by multiplying each term with 1/10 to the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 1234,123.4,12.34,1.234, _____,______ is the multiplication

Page 224 Problem 10 Answer

We are given a sequence1,−2,3,−4,5,______,_____

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence1,−2,3,−4,5,_____,____

We can observe here a pattern like adding one and then multiplying with −1 and then subtracting one and multiplying with−1i.e.

(1+1)⋅−1=−2

(−2−1)⋅−1=3

(3+1)⋅−1=−4,​so ​on

​The mathematical operations that can be used to obtain the rest terms of the sequence1,−2,3,−4,5,_____,____ are addition, subtraction and multiplication

Page 224 Problem 11 Answer

We are given a sequence−20,−16,−12,−8,−4,___,____,_____

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence−20,−16,−12,−8,−4,_____ here, we can observe a common difference of 4 i.e.−16−(−20)=4−12−(−16)=4,so on​

​From the given sequence we can observe that the value of a second term can be obtained by adding 4 with the first term.

Similarly, the other terms can be obtained by adding 4 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence−20,−16,−12,−8,−4,_____ is the addition.

Page 224 Problem 12 Answer

We are given a sequence1,10,100,1000,….,……

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation

We are given a sequence1,10,100,1000,….,……

Here, we can observe a common ratio of ​10/1

=10/100

10=10,so on

From the given sequence we can observe that the value of a second term can be obtained by multiplying 10 with the first term.

Similarly, the other terms can be obtained by multiplying 10 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 1,10,100,1000,….,…… is the multiplication.

Page 224 Problem 13 Answer

We are given a sequence−5,−5/2,−5/4,−5/8

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence−5,−5/2,−5/4,−5/8

Here, we can observe a common ratio of −5/2−5

=1/2−5/4−5/2

=1/2,so on​

From the given sequence we can observe that the value of a second term can be obtained by multiplying the first term with 1/2.

Similarly, the other terms can be obtained by multiplying the previous term with 1/2 he mathematical operation that can be used to obtain the rest terms of the sequence −5,−5/2,−5/4,−5/8 is the multiplication.

Page 224 Problem 14 Answer

We are given a sequence6.5,5,3.5,2,……………

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence6.5,5,3.5,2,……………

Here, we can observe a common difference as​6.5−5=1.5

5−3.5=1.5,​so​on

​From the given sequence we can observe that the value of a second term can be obtained by subtracting 1.5 with the first term.

Similarly, the other terms can be obtained by subtracting 1.5 with the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 6.5,5,3.5,2,…………… is the subtraction.

Page 224 Problem 15 Answer

We are given a sequence86,85,83,80,76,____,_____

We have to tell which mathematical operation can be performed in order to find the next terms of the sequence.

We will be using the mathematical operation concept to tell the required operation.

We are given a sequence 86,85,83,80,76,___,_____

We can observe here the difference between the numbers is natural numbers i.e.

86−85=1

85−83=2

83−80=3,​so​ on​

From the given sequence we can observe that the value of a second term can be obtained by subtracting a natural number from the first term.

Similarly, the other terms can be obtained by subtracting natural numbers from the previous term.

The mathematical operation that can be used to obtain the rest terms of the sequence 86,85,83,80,76,___,_____  is the subtraction.

Page 224 Problem 16 Answer

We are given a sequence as−16,4,−1,1/4.We are asked to give a mathematical expression that can be performed in order to determine the next terms of the given sequence.

For that, we will use the general formula of terms of a geometric progression that is arn−1, where a,n, and r is the first term, the number of terms, and the common difference of the g.p. respectively.

We will use the general formula of geometric progression that is arn−1 to determine the next terms of the given geometric operation−16,4,−1,1/2.

Page 224 Problem 17 Answer

We are given a sequence as1473.2,1452.7,1432.2,1411.7

We are asked to give a mathematical expression that can be performed in order to determine the next terms of the given sequence.

For that, we will use the general formula a+(n−1)d, wherea,n,and d are the first term, the number of terms, and the common difference of an arithmetic sequence respectively.

The mathematical expression a+(n−1)d can be performed to determine the next terms of the given sequence1473.2,1452.7,1432.2,1411.7

Page 224 Problem 18 Answer

We are given a sequence as √5,2,√3,√2,1,0,√−1

We are asked to give a mathematical expression that can be performed in order to determine the next terms of the given sequence.

For that, we can use a general formula √6−n where n is the number of terms in the given sequence.

The mathematical expression√6−n can be used to determine the next terms of the given sequence √5,2,√3,√2,1,0,√−1

Page 224 Problem 19 Answer

We are given a sequence as−4,12,−36,108

We are asked to give a mathematical expression that can be performed in order to determine the next terms of the given sequence.

For that, we will use a general formula arn−1 where a,n,and r are the first term, the number of terms, and the common ratio of a geometric sequence respectively.

The mathematical expression arn−1 can be used to determine the next terms of the given sequence−4,12,−36,108

Page 230 Problem 20 Answer

We are given an arithmetic sequence as 11,9,7,5 having the common difference of−2.

We are asked how would the pattern change if we take a new sequence having the same starting number as the given sequence, but its common difference is 4.

The given example started at11 and had a common difference of−2.

If the common difference is 4, the pattern would become an addition of 4,instead of an addition of−2.

The pattern would become an addition of 4,instead of an addition of−2.

Page 230 Problem 21 Answer

We are given an arithmetic sequence as11,9,7,5,………..has its common difference d=−2.

We are asked whether the given sequence remains arithmetic if we change the common difference to 4.

The given sequence will remain arithmetic as it still has a common difference d=4.

A sequence having the same starting point as the given sequence11,9,7,5………,but its common difference is 4 is also an arithmetic sequence.

Because all the sequences having a common difference between their consecutive terms are arithmetic sequences.

Page 230 Problem 22 Answer

We are given an arithmetic equation as11,9,7,5,………..has the common difference,d=−2.

We are asked to write the first five terms of a sequence having the same starting number as the given sequence, but its common difference is 4.

For that, we will use the formula An=a+(n−1)d wherea,n, and d are the first term, the number of terms, and the common difference of a sequence respectively.

We are given the first term,a=11 and the common difference,

d=4 of a sequence.

Now we will use these values in the formula An=a+(n−1)d, we get

The second term of the sequence

A2=11+(2−1)4

=11+4

=15​

The third term of the sequence is

A3=11+(3−1)4

=11+8

=19​

The fourth term of the sequence

A4=11+(4−1)4

=11+12

=23​

The fifth term of the sequence

A5=11+(5−1)4

=11+16

=27​

The first five terms of a sequence having the first term,a=11 and the common difference d=4 are11,15,19,23 and 27.

Page 230 Problem 23 Answer

We are given a set of sequences.

We are asked to list those sequences that are arithmetic.

For that, we will find the common difference of each sequence.

We are given,B=−4,−2,0,2,……. is an arithmetic sequence as it has a common difference of 2.

H=−20,−16,−12,−8,………..is an arithmetic sequence as it has a common difference of 4.

J=−5,−5/2,−5/4,−5/8,………. is an arithmetic sequence as it has a common difference of 5/2.

K=6.5,5,3.5,2,………. is an arithmetic sequence as it has a common difference of−1.5

N=1473.2,1452,7,1432.2,……… Is an arithmetic sequence as it has a common difference of−20.5.

B,H,J,K,N are the arithmetic sequences in problem 1.

Page 230 Problem 24 Answer

We are given a set of sequences.

We are asked to write the common difference of each arithmetic sequence.

For that, we will find the difference between the successive term of a sequence.

We are given B,E,H,K,N are arithmetic sequences.

Now we will find the difference between the successive term of a sequence, we get

The common difference of

Sequence B is d=2

Sequence H is  d=4

Sequence J is d=5/2

Sequence K Is d=−1.5

Sequence N is  d=−20.5.

The common difference on each arithmetic sequence B,H,J,K,N is 2,4,−5/2,−1.5,−20.5.

Page 231 Problem 25 Answer

We are given a geometric sequence as1,2,4,8….. having its common ratio r=2

We are asked how would the pattern change if we consider a sequence having the same starting number as the given sequence, but its common ratio is 3.

The given sequence started at 1 and had a common ratio of 2. If the common ratio is 3, the pattern would become multiplied by 3 instead of multiplied by 2.

The pattern change of the given sequence is1,3,9,27,…. having the common ratio is 3.

Page 231 Problem 27 Answer

We are given a geometric sequence1,2,4,8,…….

We are asked whether the new sequence has the same starting number as the sequence in the worked example, but its common ratio of 3, is still a geometric sequence.

Yes, the sequence is still geometric since the sequence still has a common ratio,

The new sequence has the common ratio 3 is still a geometric sequence.

Page 231 Problem 28 Answer

We are given a geometric sequence1,2,4,8,…….having the common ratio of 2.

We are asked to write the first 5 terms for the new sequence having the same starting number as the sequence in the worked example, but its common ratio is 3.

For that, we will use the general formula arn−1

Where a,r and n are the first term, the common ratio, and the number of terms of a geometric sequence.

We are given

The first term,a=1

The common ratio,r=3

Now we will use the formula a rn−1 to determine the first five terms of the sequence, we get

The second term of the sequence

=1(3)2−1

=(3)1

=3​

The third term of the sequence

​=1(3)3−1

=(3)2

=9​

The fourth term of the sequence

​=1(3)4−1

=(3)3

=27​

The fifth term of the sequence

=1(3)5−1

=34

=81​

The first five terms for the new sequence are : 1,3,9,27,81

Page 232 Problem 29 Answer

We have been given a sequence as the same starting number and its common ratio is 1/3

We have been asked How would the pattern change.We will use the property of sequence.

The common ratio is 1/3.

The worked out example started at 1 which means the first term is 1 and had common ratio of 2.

If the common ratio is 1/3, the pattern would become multiplied by 1/3 instead of multiplied by 2.

The pattern would change as : the pattern would become multiplied by 1/3 instead of multiplied by 2.

Page 232 Problem 30 Answer

We have been given a sequence has the same starting number and its common ratio is 1/3.

We have been asked that Is the sequence still geometric? Why or why not.As the common ratio is 1/3.

Therefore, Yes, the sequence is still geometric because the sequence still has a common ratio.

We conclude that Yes, the sequence is still geometric since the sequence still has a common ratio.

Page 232 Problem 31 Answer

We have been given a sequence has the same starting number and its common ratio is 1/3.

We have been asked the first 6 terms for the new sequence.

We will use the Sequence method.

The common ratio is 1/3.

Let, the first 6 terms for the new sequence : a1,a2,a3,a4,a5,a6

The first term is a1=1

So, the next five terms are :

a2:1⋅1/3=1/3

a3:1/3(1/3)=1/9

a4:1/9(1/3)=1/27

a5:1/27(1/3)=1/81

a6:1/81(1/3)=1/243

Hence, the new sequence is : 1,1/3,1/9,1/27,1/81,1/243

The first 6 terms of new sequence are : 1,1/3,1/9,1/27,1/81,1/243

Page 232 Problem 32 Answer

We have been given a sequence has the same starting number and its common ratio is−2.

We have been asked the how would the pattern change.

The common ratio is−2.

The worked out example started at 1 which means the first term is 1 and had a common ratio is 2.

If the common ratio is−2, the pattern would become multiply by−2 instead of multiply by 2.

The pattern would change as : the pattern would become multiplied by−2 instead of multiplied by 2.

Page 232 Problem 33 Answer

We have been given a sequence has the same starting number and its common ratio is−2.

We have been asked that Is the sequence still geometric? Explain your reasoning.

As The common ratio is−2.

Yes, the sequence is still geometric because the sequence still has a common ratio.

Yes, the sequence is still geometric because the sequence still has a common ratio.

Page 232 Problem 34 Answer

We have been given a sequence has the same starting number and its common ratio is−2

We have been asked to find the first 6 terms for the new sequence.We will use the Sequence method.

The common ratio is −2.

Let, the first 6 terms of the new sequence are : a1,a2,a3,a4,a5,a6

The first term is a1=1

So, the other five terms are :

a2:1(−2)=−2

a3:−2(−2)=4

a4:4(−2)=−8

a5:−8(−2)=16

a6:16(−2)=−32

Hence, the new sequence is :1,−2,4,−8,16,−32

The first 6 terms for the new sequence are :1,−2,4,−8,16,−32

Page 232 Problem 35 Answer

We have been given two statements are given by Devona and Chase and a sequence.

We have been asked who is correct and explain the reason.

The common ratio is the value which we multiply by to get to the next term.

Dividing by 3 is the same as multiplying by 1/3. So, the common ratio is 1/3.

Therefore, Devon a is correct.

Devon is correct as we come to conclusion that 1/3 is a common ratio.

Page 233 Problem 36 Answer

In this question, we have to analyze the sequences we cut out in Problem 1.

We have to list those sequences that are geometric.

As we know, a geometric sequence is a sequence of numbers in which the ratio of every two consecutive numbers is always a constant.

In problem 1, we cut out the sequences through A to P.

From the sequences A to P, the sequences that had to multiply or divide were A,C,F,I,J,M, and P so these sequences are geometric.

The sequences that are geometric from A to P  is A,C,F,I,J,M,and P.

Page 233 Problem 37 Answer

In this question, we have to analyze the sequences we cut out in Problem 1.

We have to write the common ratio on each geometric sequence we cut out.

As we know, a geometric sequence is a sequence of numbers in which the ratio of every two consecutive numbers is always a constant.

In problem 1, we cut out the sequences through A to P.

Sequence A was multiplying by 2 so the common ratio is r=2.

Sequence C was multiplying by 3 so the common ratio is r=3.

Sequence F was multiplying by 0.1 so the common ratio is r=0.1,

Sequence I was multiplying by 10 so the common ratio is r=10.

Sequence Jwas dividing by 2 so the common ratio is r=1/2.

Sequence M was dividing by−4 so the common ratio is r=−1/4.

Sequence P was multiplying by−3 so the common ratio is r=−3.

The common ratio in each geometric sequence A,C,F,I,J,M,P we cut out is2,3,0.1,10,1/2,−1/4,−3.

Page 233 Problem 38 Answer

In this question, we have to consider the sequences from problem 1 that are neither arithmetic nor geometric.

We have to list those sequences.As we know, a geometric sequence is a sequence of numbers in which the ratio of every two consecutive numbers is always a constant.

And we also know that the arithmetic sequence is a sequence of numbers that follow a pattern of adding a fixed number from one term to the next.

In problem 1, the sequences we are given is A to P.

The sequences that were categorized as other Pattern were D,G,L, and O.

So, these are the sequences that are neither arithmetic nor geometric.

The sequence through A to P which are neither arithmetic nor geometric is D,G,L,O.

Page 233 Problem 39 Answer

In this question, we have to consider the sequences from problem 1 that are neither arithmetic nor geometric.

We have to explain why these sequences are neither arithmetic nor geometric.

As we know, a geometric sequence is a sequence of numbers in which the ratio of every two consecutive numbers is always a constant.

And we also know that the arithmetic sequence is a sequence of numbers that follow a pattern of adding a fixed number from one term to the next.

The sequences that neither arithmetic nor geometric because they don’t have common differences and common ratios.

These sequences are not arithmetic since they do not have a common difference and they are not geometric since they do not have a common ratio.

These sequences are not arithmetic since they do not have a common difference and they are not geometric since they do not have a common ratio.

Page 234 Problem 40 Answer

We have given the first two terms of Sequence: 3,6.

Here, Dante wrote the sequence for the given terms is3,6,9,12,…And Kirs wrote the sequence for the given terms is 3,6,12,24,… We have to tell who is correct and explain our reasoning.

Only knowing the first two terms is not enough terms to determine the pattern.

The sequence was written by Dante for the given terms 3,6,9,12,… was the arithmetic sequence

If the sequence is arithmetic, then Dante sequence is correct.

The sequence was written by Kira for the given terms 3,6,12,24,… was the geometric sequence sequence

If the sequence is geometric, then kira is correct.

Since the pattern cannot be determined, either sequence could be possible answer.

Dante and Kira both are correct.

The Dante sequence is an arithmetic sequence and Kira sequence is a geometric sequence.

Page 234 Problem 41 Answer

We have given that we use question 9 here.We have to write a sequence that is neither arithmetic nor geometric.

As we know, An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

And In Geometric Sequence, each term is found by multiplying the previous term by a constant.

From question 9, we make a sequence from the given terms 3,6,…is 3,6,10,15,21,… which is not arithmetic.

Since it does not have a common difference and it is also not geometric.

Since it does not have a common ratio.

From the sequence, we get the pattern is add 3, add 4 , add 5, add 6,… since 3+3=6,

6+4=10,10+5=15, and 15+6=21

The sequence we take is 3,6,10,15,21,…and the pattern we get from the sequence that firstly add 3,then add 4, then add 5, then add 6,… and so on.

Page 234 Problem 42 Answer

In this question, we have to tell how many terms our partner needs before the pattern was recognized.

Here we use question 10.From question 10, the sequence we take there is 3,6,10,15,21,….Thus, my partner needed to know the first four terms to find the pattern.

So he/she could see the terms were going up by adding 3, then 4, and then 5.

My partner needed 4 terms before the pattern was recognized.

Page 234 Problem 43 Answer

In this question, the given sequence is2,2,2,2,2,……We have to identify the type of sequence it is and describe the pattern.

As we know, patterns are a set of numbers arranged in a sequence such that they are related to each other in a specific rule.

For the given sequence2,2,2,2,2,….

When2+0=2 the sequence could be arithmetic with a common difference of 0.

When2(1)=2 so the sequence could also be a geometric sequence with a common ratio of 1.

The given sequence 2,2,2,2,… is an arithmetic sequence when d=0 and geometric sequence when r=1.

 

Envision Math Grade 8 Volume 1 Chapter 3 Use Functions To Model Relationships

Page 165 Exercise 1 Answer

Given

Day inputs 1,2,3,4,5,6,7

Distance 30,28,30,27,30,24,36

How long the whale cross the 3,100 miles?

To find out how long will it take for the whale to travel 3,100 miles to their breeding grounds first we have to find the average distance the whale travels in one day.

We do this by adding all the values and after that dividing by the number of values that we added.

30 + 28 + 30 + 27 + 30 + 24 + 36 = 205

Now simply divide the value with 7

205 ÷ 7 = 29.286

Now that we have found how much the whale travels daily we can simply divide the number of miles needed to cross with his daily momentum,

3,100 ÷ 29.286 = 105.85

Since the real life values can not be decimal numbers, the whale needs about 106 days to travel to breeding grounds.

The whale needs about 106 days to travel 3,100 miles.

 

Page 166 Question 1 Answer

Here we have to explain the different representations of a function.

Definition and Symbolic Representation
Quadratic functions can be represented symbolically by the equation, y(x) = ax² + bx + c, where a, b, and care constants, and a ≠ 0.

This form is referred to as the standard form.

A function can be represented in three ways,

Graph

Table of data

Equation

The graph, table of data, or an equation are the three different representations of a function.

 

Page 166 Exercise 1 Answer

Given that, the total volume of the swimming pool is 9000 gallons.

As the pump is pumping water, the amount of water in the pool decreases at a constant rate. We need to graph the function.

We need to find the amount of water remaining in the pool, the amount of water pumped each hour, and the corresponding linear equation.

The total volume of the swimming pool is 9000 gallons.

From the given graph, the pump is pumping water at 750 gallons per hour.

The start of the graph is the y−intercept.

Thus, y−intercept = 9000

Here the equation corresponding to the given scenario is,

y = 9000 − 750x

where x is the number of hours and y is the number of water remaining after x hours.

The graph is,

The amount of water remaining in the pool is 9000 gallons.

The amount of water pumped each hour is 750 gallons.

The equation is y = 9000 − 750x.

The graph is,

 

Page 167 Exercise 2 Answer

We need to draw a graph that represents a linear function. Also, find an equation that represents the function.

When we graph linear functions, it is shown as a straight line in the graph.

The form of linear equations will be

y = mx + b

where m is the slope and b is the y-intercept.

Here, we can draw any linear function on the graph.

Let the linear function be y = 3x − 8

The graph will be,

The graph of y = 3x – 8 will be,

 

Page 168 Exercise 1 Answer

Here we have to explain the different representations of a function.

Definition and Symbolic Representation
Quadratic functions can be represented symbolically by the equation,y(x) = ax2 + bx + c, where a,b, and c are constants, and a ≠ 0.

This form is referred to as standard form.

A function can be represented in three ways,

Graph

Table of data

Equation

The graph, table of data, or an equation are the three different representations of a function.

 

Page 168 Exercise 3 Answer

We have to explain among the ordered pairs of a function which is to be connected by a straight line or a curve on a graph.

A function’s ordered pairs can be connected by either the straight line or by the curve depending upon the type of the function.

If the function is linear, then the graph must be connected by a straight line.

If the function is quadratic, then the graph must be connected by a curve.

Thus, we can use both depending upon the type of the function.

The ordered pairs of a function can be connected by either a straight line or a curve on a graph depending upon the type of function the ordered pairs represent.

 

Page 168 Exercise 4 Answer

Given that, Darlene tracks the number of party hats her company has in stock. The table shows the weekly stock. We need to find whether the relationship is a linear function or not using the table given.

Plotting the values in the table in a graph, we get,

The graph obtained represents a straight line.

Hence, it is linear.

The given relationship is a linear function.

The graph obtained by plotting the given values is,

 

Page 169 Exercise 6 Answer

We need to determine whether the given graph represents a function or not.

The given graph is,

The above graph represents a straight line.

Thus, it represents a linear function.

This is because the function represented by a straight line is a linear function.

The given graph represents a linear function.

 

Page 169 Exercise 7 Answer

We need to determine whether the given graph represents a function or not.

The graph is,

From the above graph, the line is not a straight line rather it is a curve.

Check if there are multiple x coordinates for every y coordinates given.

Here, there is only one i.e., an unique x coordinate for every y coordinate.

Thus, the given relationship is a function.

The given graph represents a function.

 

Page 169 Exercise 9 Answer

Given that, the relationship between the number of hexagons ,x and the perimeter of the figure they form ,y is shown in the graph. We need to find whether the perimeter of the figure a function of the number of hexagons. The graph is,

Connecting all the points together in the given graph, we get,

The above graph represents a straight line.

Thus, it represents a linear function.

This is because the function represented by a straight line is a linear function.

The perimeter of the figure is a function of the number of hexagons.

 

Page 169 Exercise 10 Answer

We need to determine whether the ordered pairs plotted in the graph below represent a function or not.

From the above graph, the line is not a straight line.

Therefore, check if there are multiple x coordinates for every y coordinates given.

Here, there are more than one x coordinate which corresponds to the same y coordinate.

From the above graph, each x value does not correspond to only one y value.

The graph does not represent a function.

The ordered pairs plotted in the graph don’t represent a function.

 

Page 170 Exercise 11 Answer

Given that, A train leaves the station at the time t=0. Traveling at a constant speed, the train travels 360 kilometers in 3 hours.

We need to write a function that relates the distance traveled, d to the timet.

Finding the distance travelled by the train in one hour, thus we get,

\(\frac{360}{3}=\frac{120 k m}{h r}\)

This implies that the train travels at a constant rate of 120 km/hr.

Therefore, the equation corresponding to this will be,

d = 120t

where t is the number of hours and d is the distance travelled.

The function that relates the distance traveled d to the time t is d = 120t

Given that, A train leaves the station at time t = 0. Traveling at a constant speed, the train travels 360 kilometers in 3hours.

We need to graph the function and tell whether it is a linear function or a nonlinear function.

The function that relates the distance traveled d to the time t is given by

d = 120t

Graphing the obtained function, we get,

The above graph represents a straight line.

Thus, it represents a linear function.

This is because the function represented by a straight line is a linear function.

From, the graph, the distance is proportional to the time.

The given function is a linear function.

The graph of the given function is,

 

Page 170 Exercise 12 Answer

We need to check and tell whether each graph is a function and justify your answer. Also, determine which graph is not a good representation of a real-world situation. The given graphs are,

The given graph is,

The above graphs doesn’t represent a straight line.

Therefore, check if there are multiple x coordinates for every y coordinates given.

In both Graph A and Graph B’s, there are more than one x coordinate which corresponds to the same y coordinate.

Thus, both of them doesn’t represent a function.

Also, in Graph A, the graph rises to the left and coming back to the right again.

This refers that the time goes back downwards. This is not possible in real life.

The graphs do not represent a function and graph A does not represent the real-world situation.

 

Page 170 Exercise 13 Answer

Given that, You have an ant farm with 22 ants. The population of ants on your farm doubles every 3 months. We need to complete the table. Also, find whether the relation is a function or not. If so, determine if it is a linear function or a nonlinear function.

Here, we have to determine the graph represents the function or not.

According to the question the ant farm with 22 ants. The population of the ants on the farm doubles every three months.

Thus, the table will become,

Plotting the ordered pairs obtained in a graph, we get,

From the above graph, the line is not a straight line.

Therefore, check if there are multiple x coordinates for every y coordinates given.

Here, there is only one x coordinate which corresponds to one y coordinate.

Therefore, the function is non-linear.

The relation is a non-linear function.

 

Page 170 Exercise 14 Answer

Given:

\(y=\frac{3}{2} x+3\)

Find: fill the table

Therefore the required solution is: