Chapter 1, Problem 52RE

a.

To determine

To find:approximate of enrolment from $2005$ through $2012$ by using graphing utility table feature.

Answer to Problem 52RE

t	$n\left(t\right)$
6	63.8
7	64.9
8	65.2
9	67.75
10	68
11	68.35
12	68

Explanation of Solution

Given information :

The numbers of students enrolled $n\left(t\right)=\left\{\begin{array}{l}-0.35t^3+6.95t^2-44.8t+158,5\le t\le 8\\ -0.35t^2+7.7t+26,8<t\le 12\end{array}\right\}$.

For $2005$ , $t=5$.

Calculation : since for $2005$ , $t=5$ then substitute the value of t in the function,

t	$n\left(t\right)$
6	63.8
7	64.9
8	65.2
9	67.75
10	68
11	68.35
12	68

b.

To determine

To graph: the modal using graphing utility and estimate the enrolment for the years $2013$ through $2017$.

Explanation of Solution

Given information :

The numbers of students enrolled $n\left(t\right)=\left\{\begin{array}{l}-0.35t^3+6.95t^2-44.8t+158,5\le t\le 8\\ -0.35t^2+7.7t+26,8<t\le 12\end{array}\right\}$.

For $2005$ , $t=5$.

Graph : by graphing utility the table for graph can be following as:

t	$n\left(t\right)$
6	63.8
7	64.9
8	65.2
9	67.75
10	68
11	68.35
12	68

Thus, the graph can be obtained as,

  

Interpretation: form the above graph it can be observed that the approximate value for $2013$ through $2017$ is $66.95,65.26,62.75,59.6,55.75$. The values does not seems reasonable because first the numbers of students increases then rapidly decreases.