EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Concept explainers

Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as p…
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Question
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Chapter 3, Problem 24PS

a.

To determine

  Use a graphing utility to create a scatter plot of the data.

a.

Expert Solution
Check Mark

Answer to Problem 24PS

  EBK PRECALCULUS W/LIMITS, Chapter 3, Problem 24PS , additional homework tip 1

Explanation of Solution

Calculation:

Consider the following data ,

  (1,4.4)(1.5,4.7)(2,5.5)(4,9.9)(6,18.1)(8,33.0)

Now draw the scatter plot,

  EBK PRECALCULUS W/LIMITS, Chapter 3, Problem 24PS , additional homework tip 2

Hence, the scatter plot is drawn.

b.

To determine

Verify that the data could best be modelled by a linear model, an exponential model, or a logarithmic model,

b.

Expert Solution
Check Mark

Answer to Problem 24PS

  Exponential Model

Explanation of Solution

Given information:

Decide whether the data could best be modelled by a linear model, an exponential model, or a logarithmic model,

  (1,4.4)(1.5,4.7)(2,5.5)(4,9.9)(6,18.1)(8,33.0)

Calculation:

Consider the following data ,

  (1,4.4)(1.5,4.7)(2,5.5)(4,9.9)(6,18.1)(8,33.0)

The best model for this data is exponential model.

Hence, the best model is Exponential Model .

c.

To determine

Explain the model is an exponential model.

c.

Expert Solution
Check Mark

Answer to Problem 24PS

Points are growing rapidly,non-linearly.

Explanation of Solution

Calculation:

Consider the following data ,

  (1,4.4)(1.5,4.7)(2,5.5)(4,9.9)(6,18.1)(8,33.0)

The best model for this data is exponential model.

It is observed from the graph that points are increasing non-linearly, thus we cannot use a linear model for this condition, it is also that points are growing rapidly thus exponential model is best suited .

Hence, we choose exponential model.

d.

To determine

To find the model is an exponential model for the data and graph the model with the scatter plot.

d.

Expert Solution
Check Mark

Answer to Problem 24PS

  EBK PRECALCULUS W/LIMITS, Chapter 3, Problem 24PS , additional homework tip 3

Explanation of Solution

Calculation:

Consider the following data ,

  (1,4.4)(1.5,4.7)(2,5.5)(4,9.9)(6,18.1)(8,33.0)

By using the value of x and y from the data and graphing utility, we get

  >X:=Vector([1,1.5,2,4,6,8],datatype=float)

  X:=11.500000000000000002468

  >Y:=Vector([4.4,4.7,5.5,9.9,18.1,33],datatype=float)Y:=4.400000000000000344.700000000000000165.500000000000000169.9000000000000003418.1000000000000001433

  >Fit(ab',X,Y,t)3.03932530280103119.3471051913653324

Hence the required model is ,

  y=30114(1.341)x

Now scatter plot for this model,

  EBK PRECALCULUS W/LIMITS, Chapter 3, Problem 24PS , additional homework tip 4

Hence, the required model is y=30114(1.341)x .

e.

To determine

Describe the model you chose fits the data.

e.

Expert Solution
Check Mark

Answer to Problem 24PS

Model fits the data.

Explanation of Solution

Calculation:

Consider the required model is ,

  y=30114(1.341)x

Now scatter plot for this model,

  EBK PRECALCULUS W/LIMITS, Chapter 3, Problem 24PS , additional homework tip 5

The graph shows model fits the data,

Hence, the model fits the data.

Previous
Chapter 3, Problem 23PS
Next
Chapter 3, Problem 25PS

Chapter 3 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10E
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Prob. 1ECh. 3.4 - Prob. 2ECh. 3.4 - Prob. 3ECh. 3.4 - Prob. 4ECh. 3.4 - Prob. 5ECh. 3.4 - Prob. 6ECh. 3.4 - Prob. 7ECh. 3.4 - Prob. 8ECh. 3.4 - Prob. 9ECh. 3.4 - Prob. 10ECh. 3.4 - Prob. 11ECh. 3.4 - Prob. 12ECh. 3.4 - Prob. 13ECh. 3.4 - Prob. 14ECh. 3.4 - Prob. 15ECh. 3.4 - Prob. 16ECh. 3.4 - Prob. 17ECh. 3.4 - Prob. 18ECh. 3.4 - Prob. 19ECh. 3.4 - Prob. 20ECh. 3.4 - Prob. 21ECh. 3.4 - Prob. 22ECh. 3.4 - Prob. 23ECh. 3.4 - Prob. 24ECh. 3.4 - Prob. 25ECh. 3.4 - Prob. 26ECh. 3.4 - Prob. 27ECh. 3.4 - Prob. 28ECh. 3.4 - Prob. 29ECh. 3.4 - Prob. 30ECh. 3.4 - Prob. 31ECh. 3.4 - Prob. 32ECh. 3.4 - Prob. 33ECh. 3.4 - Prob. 34ECh. 3.4 - Prob. 35ECh. 3.4 - Prob. 36ECh. 3.4 - Prob. 37ECh. 3.4 - Prob. 38ECh. 3.4 - Prob. 39ECh. 3.4 - Prob. 40ECh. 3.4 - Prob. 41ECh. 3.4 - Prob. 42ECh. 3.4 - Prob. 43ECh. 3.4 - Prob. 44ECh. 3.4 - Prob. 45ECh. 3.4 - Prob. 46ECh. 3.4 - Prob. 47ECh. 3.4 - Prob. 48ECh. 3.4 - Prob. 49ECh. 3.4 - Prob. 50ECh. 3.4 - Prob. 51ECh. 3.4 - Prob. 52ECh. 3.4 - Prob. 53ECh. 3.4 - Prob. 54ECh. 3.4 - 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