Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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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 1.1, Problem 96E

a.

To determine

Calculate the linear equation of the line giving the demand x in terms of the rent p .

a.

Expert Solution
Check Mark

Answer to Problem 96E

The linear equation of the line is x=50−(p−58015) .

Explanation of Solution

Given:

It is given in the question that a real state office handles an apartment complex with 50 units.The rent per unit is $580 per month ,all 50 units are accupied.However,when the rent is $625 per month,the average number of accupied units drops to 47 .Also assume that the relationship between the monthly rent p and the demand x is linear.

Concept Used:

In thise,use the concept of umderstanding the different variable and constant to make the linear equation and also use the slope formula i.e, m=y2−y1x2−x1 .

Calculation: In this ,first it has to be find the slope for the linear equation .now,looking at the two points it gives us, (580,50) and (625,47) so,to find the slope we need,

  m=y2−y1x2−x1=47−50625−580=−345=−115

So by the given slope ,it means that for every 15 dollars it add onto the total rent (p)mit lose one occupied housing unit (x).

Now,it’s time to create a equation but apart from that it has to keep in mind that 50 is the maximum amount of housing units available ,therefore,it can deduce that 580 is the lowest rent that can reasonably go to maintain full occupancy,Factoring these deductions into the problem,as well as including the slope,the equation that get is ,

  x=50−(p−58015) ,where x is the number of units in demand and p is the rent charged , as laid forth in this.

Conclusion:

The equation is x=50−(p−58015)

b.

To determine

Draw the graph of the equation obtained in part (a) and also Find the number of units occupied when the rent is $655 .

b.

Expert Solution
Check Mark

Answer to Problem 96E

The number of units occupied is 45 units.

Explanation of Solution

Given:

It is given in the question that the expression obtained from part (a) is x=50−(p−58015) .

Concept Used:

In this, use the concept of graphical analysis and use the expression to put the value on that and find the solution.

Calculation:

The graph of the equation is given below:

  Precalculus with Limits: A Graphing Approach, Chapter 1.1, Problem 96E , additional homework tip 1

Now,the equation is x=50−(p−58015).

Put p=655 in the equation above,

  x=50−(p−58015)x=50−(655−58015)x=50−(7515)x=50−5=45

Conclusion:

The number of unit occupied is 45 units.

c.

To determine

Find the number of units occupied when the rent is lowered to $595 ,by using equation verify graphically.

c.

Expert Solution
Check Mark

Answer to Problem 96E

The number of units accupied is 49 units.

Explanation of Solution

Given:

It is given in the question that the expression is x=50−(p−58015) .

Concept Used:

In this, use the concept of putting the value in the expression and find the solution and also known to draw the graph.

Calculation: Now,the equation is x=50−(p−58015).

Put p=595 in the equation above,

  x=50−(p−58015)x=50−(595−58015)x=50−(1515)x=50−1=49

The graph for this is also given below:

  Precalculus with Limits: A Graphing Approach, Chapter 1.1, Problem 96E , additional homework tip 2

Conclusion:

The number of units occupied is 49 units.

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Chapter 1.1, Problem 95E
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Chapter 1.1, Problem 97E

Chapter 1 Solutions

Precalculus with Limits: A Graphing Approach

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Prob. 15ECh. 1.6 - Prob. 16ECh. 1.6 - Prob. 17ECh. 1.6 - Prob. 18ECh. 1.6 - Prob. 19ECh. 1.6 - Prob. 20ECh. 1.6 - Prob. 21ECh. 1.6 - Prob. 22ECh. 1.6 - Prob. 23ECh. 1.6 - Prob. 24ECh. 1.6 - Prob. 25ECh. 1.6 - Prob. 26ECh. 1.6 - Prob. 27ECh. 1.6 - Prob. 28ECh. 1.6 - Prob. 29ECh. 1.6 - Prob. 30ECh. 1.6 - Prob. 31ECh. 1.6 - Prob. 32ECh. 1.6 - Prob. 33ECh. 1.6 - Prob. 34ECh. 1.6 - Prob. 35ECh. 1.6 - Prob. 36ECh. 1.6 - Prob. 37ECh. 1.6 - Prob. 38ECh. 1.6 - Prob. 39ECh. 1.6 - Prob. 40ECh. 1.6 - Prob. 41ECh. 1.6 - Prob. 42ECh. 1.6 - Prob. 43ECh. 1.6 - Prob. 44ECh. 1.6 - Prob. 45ECh. 1.6 - Prob. 46ECh. 1.6 - Prob. 47ECh. 1.6 - Prob. 48ECh. 1.6 - Prob. 49ECh. 1.6 - Prob. 50ECh. 1.6 - Prob. 51ECh. 1.6 - Prob. 52ECh. 1.6 - Prob. 53ECh. 1.6 - Prob. 54ECh. 1.6 - Prob. 55ECh. 1.6 - Prob. 56ECh. 1.6 - Prob. 57ECh. 1.6 - Prob. 58ECh. 1.6 - Prob. 59ECh. 1.6 - Prob. 60ECh. 1.6 - Prob. 61ECh. 1.6 - Prob. 62ECh. 1.6 - Prob. 63ECh. 1.6 - Prob. 64ECh. 1.6 - Prob. 65ECh. 1.6 - Prob. 66ECh. 1.6 - Prob. 67ECh. 1.6 - Prob. 68ECh. 1.6 - Prob. 69ECh. 1.6 - Prob. 70ECh. 1.6 - Prob. 71ECh. 1.6 - Prob. 72ECh. 1.6 - Prob. 73ECh. 1.6 - Prob. 74ECh. 1.6 - Prob. 75ECh. 1.6 - Prob. 76ECh. 1.6 - Prob. 77ECh. 1.6 - Prob. 78ECh. 1.6 - Prob. 79ECh. 1.6 - Prob. 80ECh. 1.6 - Prob. 81ECh. 1.6 - Prob. 82ECh. 1.6 - Prob. 83ECh. 1.6 - Prob. 84ECh. 1.6 - Prob. 85ECh. 1.6 - Prob. 86ECh. 1.6 - Prob. 87ECh. 1.6 - Prob. 88ECh. 1.6 - Prob. 89ECh. 1.6 - Prob. 90ECh. 1.6 - Prob. 91ECh. 1.6 - Prob. 92ECh. 1.6 - Prob. 93ECh. 1.6 - Prob. 94ECh. 1.6 - Prob. 95ECh. 1.6 - Prob. 96ECh. 1.6 - Prob. 97ECh. 1.6 - Prob. 98ECh. 1.6 - Prob. 99ECh. 1.6 - Prob. 100ECh. 1.6 - Prob. 101ECh. 1.6 - Prob. 102ECh. 1.6 - Prob. 103ECh. 1.6 - Prob. 104ECh. 1.6 - Prob. 105ECh. 1.6 - Prob. 106ECh. 1.6 - Prob. 107ECh. 1.6 - Prob. 108ECh. 1.6 - Prob. 109ECh. 1.6 - Prob. 110ECh. 1.6 - Prob. 111ECh. 1.6 - Prob. 112ECh. 1.6 - Prob. 113ECh. 1.6 - Prob. 114ECh. 1.6 - Prob. 115ECh. 1.6 - Prob. 116ECh. 1.6 - Prob. 117ECh. 1.6 - Prob. 118ECh. 1.6 - Prob. 119ECh. 1.6 - Prob. 120ECh. 1.6 - Prob. 121ECh. 1.6 - Prob. 122ECh. 1.6 - Prob. 123ECh. 1.6 - Prob. 124ECh. 1.6 - Prob. 125ECh. 1.6 - Prob. 126ECh. 1.6 - Prob. 127ECh. 1.6 - Prob. 128ECh. 1.6 - Prob. 129ECh. 1.6 - Prob. 130ECh. 1.6 - Prob. 131ECh. 1.6 - Prob. 132ECh. 1.6 - Prob. 133ECh. 1.6 - Prob. 134ECh. 1.6 - Prob. 135ECh. 1.6 - Prob. 136ECh. 1.6 - Prob. 137ECh. 1.6 - Prob. 138ECh. 1.6 - Prob. 139ECh. 1.6 - Prob. 140ECh. 1.7 - Prob. 1ECh. 1.7 - Prob. 2ECh. 1.7 - Prob. 3ECh. 1.7 - Prob. 4ECh. 1.7 - Prob. 5ECh. 1.7 - Prob. 6ECh. 1.7 - Prob. 7ECh. 1.7 - Prob. 8ECh. 1.7 - Prob. 9ECh. 1.7 - Prob. 10ECh. 1.7 - Prob. 11ECh. 1.7 - Prob. 12ECh. 1.7 - Prob. 13ECh. 1.7 - Prob. 14ECh. 1.7 - Prob. 15ECh. 1.7 - Prob. 16ECh. 1.7 - Prob. 17ECh. 1.7 - Prob. 18ECh. 1.7 - Prob. 19ECh. 1.7 - Prob. 20ECh. 1.7 - Prob. 21ECh. 1.7 - Prob. 22ECh. 1.7 - Prob. 23ECh. 1.7 - Prob. 24ECh. 1.7 - Prob. 25ECh. 1.7 - Prob. 26ECh. 1.7 - Prob. 27ECh. 1.7 - Prob. 28ECh. 1.7 - Prob. 29ECh. 1.7 - Prob. 30ECh. 1.7 - Prob. 31ECh. 1.7 - Prob. 32ECh. 1.7 - Prob. 33ECh. 1.7 - Prob. 34ECh. 1.7 - Prob. 35ECh. 1.7 - Prob. 36ECh. 1.7 - Prob. 37ECh. 1.7 - Prob. 38ECh. 1 - Prob. 1RECh. 1 - Prob. 2RECh. 1 - Prob. 3RECh. 1 - Prob. 4RECh. 1 - Prob. 5RECh. 1 - Prob. 6RECh. 1 - Prob. 7RECh. 1 - Prob. 8RECh. 1 - Prob. 9RECh. 1 - Prob. 10RECh. 1 - Prob. 11RECh. 1 - Prob. 12RECh. 1 - Prob. 13RECh. 1 - Prob. 14RECh. 1 - Prob. 15RECh. 1 - Prob. 16RECh. 1 - Prob. 17RECh. 1 - Prob. 18RECh. 1 - Prob. 19RECh. 1 - Prob. 20RECh. 1 - Prob. 21RECh. 1 - Prob. 22RECh. 1 - Prob. 23RECh. 1 - Prob. 24RECh. 1 - Prob. 25RECh. 1 - Prob. 26RECh. 1 - Prob. 27RECh. 1 - Prob. 28RECh. 1 - Prob. 29RECh. 1 - Prob. 30RECh. 1 - Prob. 31RECh. 1 - Prob. 32RECh. 1 - Prob. 33RECh. 1 - Prob. 34RECh. 1 - Prob. 35RECh. 1 - Prob. 36RECh. 1 - Prob. 37RECh. 1 - Prob. 38RECh. 1 - Prob. 39RECh. 1 - Prob. 40RECh. 1 - Prob. 41RECh. 1 - Prob. 42RECh. 1 - Prob. 43RECh. 1 - Prob. 44RECh. 1 - Prob. 45RECh. 1 - Prob. 46RECh. 1 - Prob. 47RECh. 1 - Prob. 48RECh. 1 - Prob. 49RECh. 1 - Prob. 50RECh. 1 - Prob. 51RECh. 1 - Prob. 52RECh. 1 - Prob. 53RECh. 1 - Prob. 54RECh. 1 - Prob. 55RECh. 1 - Prob. 56RECh. 1 - Prob. 57RECh. 1 - Prob. 58RECh. 1 - Prob. 59RECh. 1 - Prob. 60RECh. 1 - Prob. 61RECh. 1 - Prob. 62RECh. 1 - Prob. 63RECh. 1 - Prob. 64RECh. 1 - Prob. 65RECh. 1 - Prob. 66RECh. 1 - Prob. 67RECh. 1 - Prob. 68RECh. 1 - Prob. 69RECh. 1 - Prob. 70RECh. 1 - Prob. 71RECh. 1 - Prob. 72RECh. 1 - Prob. 73RECh. 1 - Prob. 74RECh. 1 - Prob. 75RECh. 1 - Prob. 76RECh. 1 - Prob. 77RECh. 1 - Prob. 78RECh. 1 - Prob. 79RECh. 1 - Prob. 80RECh. 1 - Prob. 81RECh. 1 - Prob. 82RECh. 1 - Prob. 83RECh. 1 - Prob. 84RECh. 1 - Prob. 85RECh. 1 - Prob. 86RECh. 1 - Prob. 87RECh. 1 - Prob. 88RECh. 1 - Prob. 89RECh. 1 - Prob. 90RECh. 1 - Prob. 91RECh. 1 - Prob. 92RECh. 1 - Prob. 93RECh. 1 - Prob. 94RECh. 1 - Prob. 95RECh. 1 - Prob. 96RECh. 1 - Prob. 97RECh. 1 - Prob. 98RECh. 1 - Prob. 99RECh. 1 - Prob. 100RECh. 1 - Prob. 101RECh. 1 - Prob. 102RECh. 1 - Prob. 103RECh. 1 - Prob. 104RECh. 1 - Prob. 105RECh. 1 - Prob. 106RECh. 1 - Prob. 107RECh. 1 - Prob. 108RECh. 1 - Prob. 109RECh. 1 - Prob. 110RECh. 1 - Prob. 111RECh. 1 - Prob. 112RECh. 1 - Prob. 113RECh. 1 - Prob. 114RECh. 1 - Prob. 115RECh. 1 - Prob. 116RECh. 1 - Prob. 117RECh. 1 - Prob. 118RECh. 1 - Prob. 119RECh. 1 - Prob. 120RECh. 1 - Prob. 121RECh. 1 - Prob. 122RECh. 1 - Prob. 123RECh. 1 - Prob. 124RECh. 1 - Prob. 125RECh. 1 - Prob. 126RECh. 1 - Prob. 127RECh. 1 - Prob. 128RECh. 1 - Prob. 129RECh. 1 - Prob. 130RECh. 1 - Prob. 131RECh. 1 - Prob. 132RECh. 1 - Prob. 133RECh. 1 - Prob. 134RECh. 1 - Prob. 135RECh. 1 - Prob. 136RECh. 1 - Prob. 137RECh. 1 - Prob. 138RECh. 1 - Prob. 139RECh. 1 - Prob. 140RECh. 1 - Prob. 141RECh. 1 - Prob. 142RECh. 1 - Prob. 143RECh. 1 - Prob. 144RECh. 1 - Prob. 145RECh. 1 - Prob. 146RECh. 1 - Prob. 147RECh. 1 - Prob. 148RECh. 1 - Prob. 149RECh. 1 - Prob. 150RECh. 1 - Prob. 151RECh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 13CTCh. 1 - Prob. 14CTCh. 1 - Prob. 15CTCh. 1 - Prob. 16CTCh. 1 - Prob. 17CTCh. 1 - Prob. 18CTCh. 1 - Prob. 19CTCh. 1 - Prob. 20CT
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