Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Review Exercises - Page 513: 48

Answer

a) The simplified value is $76$. b) The simplified average score is $67,\text{ 63, 61, 59, and 56}$. c) Shown below
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Work Step by Step

(a) For the first exam, the value of t is 0: $\begin{align} & f\left( 0 \right)=76-18\log \left( 0+1 \right) \\ & =76-18\log \left( 1 \right) \\ & =76-18\left( 0 \right) \\ & =76 \end{align}$ Thus, the average score for the first exam is 76. (b) We have the average score after the time period of 2 months: $\begin{align} & f\left( 2 \right)=76-18\log \left( 2+1 \right) \\ & =76-18\log \left( 3 \right) \\ & =76-18\left( 0.4771 \right) \\ & =67.4118 \end{align}$ Take the approximate value: $ f\left( 2 \right)\approx 67$ And the average score after the time period of 4 months is: $\begin{align} & f\left( 4 \right)=76-18\log \left( 4+1 \right) \\ & =76-18\log \left( 5 \right) \\ & =76-18\left( 0.6990 \right) \\ & =63.4185 \end{align}$ Take the approximate value: $ f\left( 4 \right)\approx 63$ And the average score after the time period of 6 months is: $\begin{align} & f\left( 6 \right)=76-18\log \left( 6+1 \right) \\ & =76-18\log \left( 7 \right) \\ & =76-18\left( 0.8451 \right) \\ & =60.7882 \end{align}$ Take the approximate value: $ f\left( 6 \right)\approx 61$ The average score after the time period of 8 months is: $\begin{align} & f\left( 8 \right)=76-18\log \left( 8+1 \right) \\ & =76-18\log \left( 9 \right) \\ & =76-18\left( 0.9542 \right) \\ & =58.8236 \end{align}$ Take the approximate value: $ f\left( 8 \right)\approx 59$ The average score after the time period of 1 year or 12 months is: $\begin{align} & f\left( 12 \right)=76-18\log \left( 12+1 \right) \\ & =76-18\log \left( 13 \right) \\ & =76-18\left( 1.1139 \right) \\ & =55.9490 \end{align}$ Take the approximate value: $ f\left( 12 \right)\approx 56$ Thus, the average scores after 2 months, 4 months, 6 months, 8 months, and 1 year are $67,\text{ 63, 61, 59, and 56}$ respectively. (c) Draw the graph, formulate the table from values of t and f obtained in part(a) and part(b). Thus, as the value of t increases, the value of $76-18\log \left( t+1 \right)$ decreases. We see that with an increase in time t, the retention of course content decreases for students.
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