Chapter 7.3, Problem 50PPS

(a)

To determine

To Find: The student score at the time the school year ended $t=0$ ?

Answer to Problem 50PPS

  $y(0)=85$

Explanation of Solution

Given: To measure a student’s retention of knowledge, the student is tested after a given amount of time. A student score on a Algebra 2 test $t$ months after the year is over can be approximate by $y(t)=85-6{\log }_2(t+1)$ where $y(t)$ is the student score as a percent.

Given: $y(t)=85-6{\log }_2(t+1)$

When $t=0$

  $\begin{array}{l}y(t)=85-6{\log }_2(t+1)\\ y(0)=85-6{\log }_2(0+1)\\ y(0)=85-6{\log }_{2}1\\ y(0)=85-6{\log }_2{2}^0\\ y(0)=85-0{\log }_{2}2\\ y(0)=85-0\\ y(0)=85\end{array}$

Hence, $y(0)=85$

(b)

To determine

To Find: The student score after $3$ months.

Answer to Problem 50PPS

  $y(3)=73$

Explanation of Solution

Given: To measure a student’s retention of knowledge, the student is tested after a given amount of time. A student score on a Algebra 2 test $t$ months after the year is over can be approximate by $y(t)=85-6{\log }_2(t+1)$ where $y(t)$ is the student score as a percent.

Given: $y(t)=85-6{\log }_2(t+1)$

When $t=3$

  $\begin{array}{l}y(t)=85-6{\log }_2(t+1)\\ y(3)=85-6{\log }_2(3+1)\\ y(3)=85-6{\log }_{2}4\\ y(3)=85-6{\log }_2{2}^2\\ y(3)=85-12{\log }_{2}2\\ y(3)=85-12\\ y(3)=73\end{array}$

Hence, $y(3)=73$

(c)

To determine

To Find: The student score after $15$ months.

Answer to Problem 50PPS

  $y(15)=61$

Explanation of Solution

Given: To measure a student’s retention of knowledge, the student is tested after a given amount of time. A student score on a Algebra 2 test $t$ months after the year is over can be approximate by $y(t)=85-6{\log }_2(t+1)$ where $y(t)$ is the student score as a percent.

Given: $y(t)=85-6{\log }_2(t+1)$

When $t=15$

  $\begin{array}{l}y(t)=85-6{\log }_2(t+1)\\ y(15)=85-6{\log }_2(15+1)\\ y(15)=85-6{\log }_{2}16\\ y(15)=85-6{\log }_2{2}^4\\ y(15)=85-24{\log }_{2}2\\ y(15)=85-24\\ y(15)=61\end{array}$

Hence, $y(15)=61$