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

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 2.6, Problem 69E

a.

To determine

Draw the graph of the function.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

It is given in the question that the function is C=25000p100−p .

Graph:

  EBK PRECALCULUS W/LIMITS, Chapter 2.6, Problem 69E

Interpretation: Here, the function is C=25000p100−p ,put these function in the graphing calculator, it will give the required graph in proper way.

b.

To determine

Calculate the cost of supplying bins to 15%,50%, and 90% of the population.

b.

Expert Solution
Check Mark

Answer to Problem 69E

The cost of the values are 4411.7,25000 and 225000.

Explanation of Solution

Given:

It is given in the question that the function is C=25000p100−p .

Concept Used:

In this, use the concept of putting the given value in the function and get the result.

Calculation: Here, the function is C=25000p100−p .

Put, p=15%

  C=25000p100−p=25000×15100−15=37500085=4411.7

Put, p=50%

  C=25000p100−p=25000×50100−50=125000050=25000

Put, p=90%

  C=25000p100−p=25000×90100−90=225000010=225000

Conclusion:

The values are 4411.7,25000 and 225000.

c.

To determine

Explain whether it is possible to supply bins to 100% of the population.

c.

Expert Solution
Check Mark

Answer to Problem 69E

It doesn’t full fill all the quantity of the bins.

Explanation of Solution

Given:

It is given in the question that the function is C=25000p100−p .

Here, if the percentage is 100% , then it is not possible to supply all the bins to all the population because of the given ration seen below:

  C=25000p100−p=25000×100100−100=25000000=∞

So, infinity defines that it doesn’t full fill all the quantity.

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Chapter 2.6, Problem 68E
Next
Chapter 2.6, Problem 70E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 14ECh. 2.7 - Prob. 15ECh. 2.7 - Prob. 16ECh. 2.7 - Prob. 17ECh. 2.7 - Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.7 - Prob. 47ECh. 2.7 - Prob. 48ECh. 2.7 - Prob. 49ECh. 2.7 - Prob. 50ECh. 2.7 - Prob. 51ECh. 2.7 - Prob. 52ECh. 2.7 - Prob. 53ECh. 2.7 - Prob. 54ECh. 2.7 - Prob. 55ECh. 2.7 - Prob. 56ECh. 2.7 - Prob. 57ECh. 2.7 - Prob. 58ECh. 2.7 - Prob. 59ECh. 2.7 - Prob. 60ECh. 2.7 - Prob. 61ECh. 2.7 - Prob. 62ECh. 2.7 - Prob. 63ECh. 2.7 - Prob. 64ECh. 2.7 - Prob. 65ECh. 2.7 - Prob. 66ECh. 2.7 - Prob. 67ECh. 2.7 - Prob. 68ECh. 2.7 - Prob. 69ECh. 2.7 - Prob. 70ECh. 2.7 - Prob. 71ECh. 2.7 - Prob. 72ECh. 2.7 - Prob. 73ECh. 2.7 - Prob. 74ECh. 2.7 - Prob. 75ECh. 2.7 - Prob. 76ECh. 2.7 - Prob. 77ECh. 2.7 - Prob. 78ECh. 2.7 - Prob. 79ECh. 2.7 - Prob. 80ECh. 2.7 - Prob. 81ECh. 2.7 - Prob. 82ECh. 2.7 - Prob. 83ECh. 2.7 - Prob. 84ECh. 2.7 - Prob. 85ECh. 2.7 - Prob. 86ECh. 2.7 - Prob. 87ECh. 2.7 - Prob. 88ECh. 2.7 - Prob. 89ECh. 2.7 - Prob. 90ECh. 2 - Prob. 1RECh. 2 - Prob. 2RECh. 2 - Prob. 3RECh. 2 - Prob. 4RECh. 2 - Prob. 5RECh. 2 - Prob. 6RECh. 2 - Prob. 7RECh. 2 - Prob. 8RECh. 2 - Prob. 9RECh. 2 - Prob. 10RECh. 2 - Prob. 11RECh. 2 - Prob. 12RECh. 2 - Prob. 13RECh. 2 - Prob. 14RECh. 2 - Prob. 15RECh. 2 - Prob. 16RECh. 2 - Prob. 17RECh. 2 - Prob. 18RECh. 2 - Prob. 19RECh. 2 - Prob. 20RECh. 2 - Prob. 21RECh. 2 - Prob. 22RECh. 2 - Prob. 23RECh. 2 - Prob. 24RECh. 2 - Prob. 25RECh. 2 - Prob. 26RECh. 2 - Prob. 27RECh. 2 - Prob. 28RECh. 2 - Prob. 29RECh. 2 - Prob. 30RECh. 2 - Prob. 31RECh. 2 - Prob. 32RECh. 2 - Prob. 33RECh. 2 - Prob. 34RECh. 2 - Prob. 35RECh. 2 - Prob. 36RECh. 2 - Prob. 37RECh. 2 - Prob. 38RECh. 2 - Prob. 39RECh. 2 - Prob. 40RECh. 2 - Prob. 41RECh. 2 - Prob. 42RECh. 2 - Prob. 43RECh. 2 - Prob. 44RECh. 2 - Prob. 45RECh. 2 - Prob. 46RECh. 2 - Prob. 47RECh. 2 - Prob. 48RECh. 2 - Prob. 49RECh. 2 - Prob. 50RECh. 2 - Prob. 51RECh. 2 - Prob. 52RECh. 2 - Prob. 53RECh. 2 - Prob. 54RECh. 2 - Prob. 55RECh. 2 - Prob. 56RECh. 2 - Prob. 57RECh. 2 - Prob. 58RECh. 2 - Prob. 59RECh. 2 - Prob. 60RECh. 2 - Prob. 61RECh. 2 - Prob. 62RECh. 2 - Prob. 63RECh. 2 - Prob. 64RECh. 2 - Prob. 65RECh. 2 - Prob. 66RECh. 2 - Prob. 67RECh. 2 - Prob. 68RECh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 1PSCh. 2 - Prob. 2PSCh. 2 - Prob. 3PSCh. 2 - Prob. 4PSCh. 2 - Prob. 5PSCh. 2 - Prob. 6PSCh. 2 - Prob. 7PSCh. 2 - Prob. 8PSCh. 2 - Prob. 9PSCh. 2 - Prob. 10PSCh. 2 - Prob. 11PSCh. 2 - Prob. 12PSCh. 2 - Prob. 13PS
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