a. Answer Explanation Given: f ( t ) = 80 − 17 log ( t + 1 ) , 0 ≤ t ≤ 12 Graph: Using graphing utility plot f ( t ) = 80 − 17 log ( t + 1 ) , 0 ≤ t ≤ 12 b. To determine To calculate: The average score on the original exam ( t = 0 ) Answer The average score on the original exam with t = 0 is 80 Explanation Given f ( t ) = 80 − 17 log ( t + 1 ) , 0 ≤ t ≤ 12 Calculation: Put t = 0 in f ( t ) = 80 − 17 log ( t + 1 ) f ( 0 ) = 80 − 17 log ( 0 + 1 ) = 80 − 17 log ( 1 ) = 80 − 17 ( 0 ) ( ∵ log ( 1 ) = 0 ) = 80 c. To determine To calculate: The average score after 4 months Answer The average score after 4 months is 68.12 Explanation Given f ( t ) = 80 − 17 log ( t + 1 ) , 0 ≤ t ≤ 12 Calculation: Put t = 4 in f ( t ) = 80 − 17 log ( t + 1 ) f ( 0 ) = 80 − 17 ( log ( 4 + 1 ) ) = 80 − 17 ( log ( 5 ) ) = 68.12 d. To determine To calculate: The average score after 10 months Answer The average score after 10 months is 62.3 Explanation Given f ( t ) = 80 − 17 log ( t + 1 ) , 0 ≤ t ≤ 12 Calculation: Put t = 10 in f ( t ) = 80 − 17 log ( t + 1 ) f ( 10 ) = 80 − 17 log ( 10 + 1 ) = 80 − 17 log ( 11 ) = 62.3

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 3.2, Problem 79E

To determine

To Graph: the model over the specific domain using a graphing utility

Expert Solution

Answer to Problem 79E

Explanation of Solution

Given: f(t)=80−17log(t+1), 0≤t≤12

Graph:

Using graphing utility plot f(t)=80−17log(t+1), 0≤t≤12

To determine

To calculate: The average score on the original exam (t=0)

Expert Solution

Answer to Problem 79E

The average score on the original exam with t=0 is 80

Explanation of Solution

Given

f(t)=80−17log(t+1), 0≤t≤12

Calculation:

Put t=0 in f(t)=80−17log(t+1)

f(0)=80−17log(0+1)=80−17log(1)=80−17(0) (∵log(1)=0)=80

To determine

To calculate: The average score after 4 months

Expert Solution

Answer to Problem 79E

The average score after 4 months is 68.12

Explanation of Solution

Given

f(t)=80−17log(t+1), 0≤t≤12

Calculation:

Put t=4 in f(t)=80−17log(t+1)

f(0)=80−17(log(4+1))=80−17(log(5))=68.12

To determine

To calculate: The average score after 10 months

Expert Solution

Answer to Problem 79E

The average score after 10 months is 62.3

Explanation of Solution

Given

f(t)=80−17log(t+1), 0≤t≤12

Calculation:

Put t=10 in f(t)=80−17log(t+1)

f(10)=80−17log(10+1)=80−17log(11)=62.3

Chapter 3.2, Problem 78E

Chapter 3.2, Problem 80E

Chapter 3 Solutions

EBK PRECALCULUS W/LIMITS

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