Precalculus with Limits
Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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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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Chapter 3, Problem 23PS

a.

To determine

Use a graphical utility to create a sactter plot of the data.

a.

Expert Solution
Check Mark

Answer to Problem 23PS

The scatter plot of the data is given as

  Precalculus with Limits, Chapter 3, Problem 23PS , additional homework tip 1

Explanation of Solution

Calculation:

The scatter plot of the data is given as

  Precalculus with Limits, Chapter 3, Problem 23PS , additional homework tip 2

Hence, the graph is shown above.

b.

To determine

Decide the model for data .

b.

Expert Solution
Check Mark

Answer to Problem 23PS

logarithmic model.

Explanation of Solution

Calculation:

The best model for these points is logarithmic model

Hence, the data could be modelled by a logarithmic model

c.

To determine

Explain the model of part (b) .

c.

Expert Solution
Check Mark

Answer to Problem 23PS

From the graph, we can see that points are increasing non-linearly, so, we cannot use a model linear model for this. We can also see that points are not growing rapidly,so, exponential model is not suitable.

Explanation of Solution

Calculation:

From the graph, we can see that points are increasing non-linearly, so, we cannot use a model linear model for this. We can also see that points are not growing rapidly,so, exponential model is not suitable.

Hence, we chose logarithmic model.

d.

To determine

find the model you chose in part (b) for the data and graph the model with scatter plot.

d.

Expert Solution
Check Mark

Answer to Problem 23PS

  y=2.151+2.7043ln(x)

  Precalculus with Limits, Chapter 3, Problem 23PS , additional homework tip 3

Explanation of Solution

Calculation:

Now, by using the values of and from the data and graphing utility we get,

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

  X=1.1.500000000000000002.4.6.8.

  Y:Vector([2,3,5,4,5.8,7,7.8],datatype=float);

  Y=2.3.500000000000000004.5.799999999999999827.7.79999999999999982

Hence, the required model is y=2.151+2.7043ln(x) .

Now, the graph of scatter plot is,

  Precalculus with Limits, Chapter 3, Problem 23PS , additional homework tip 4

Hence, the graph is shown above.

e.

To determine

Determine if the model you chose fits the data.

e.

Expert Solution
Check Mark

Answer to Problem 23PS

From the graph we can see that the model we get fits the data very well.

Explanation of Solution

Calculation:

From the graph we can see that the model we get fits the data very well.

Hence, logarithmic model fits the data very well.

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Chapter 3, Problem 22PS
Next
Chapter 3, Problem 24PS

Chapter 3 Solutions

Precalculus with 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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