Chapter 8.1, Problem 51PPS

a.

To determine

To write an equation modelling the number of students $S$ in grades 6 through 12 enrolled from the year 2006 to 2015.

Answer to Problem 51PPS

  $S=1.95t^2-1.95t+11.3$

Explanation of Solution

Given:

Total number of students in school enrolled from 2006 to 2015, $N=1.25t^2-t+7.5$ .

Number of students in Kindergarten through fifth grade, $P=0.7t^2-0.95t+3.8$ .

Concept used:

Number of studentsin grades 6 through 12 enrolled from the year 2006 to 2015 is given by:

  $S=N-P$

Calculation: Number of students in grades 6 through 12 enrolled from the year 2006 to 2015 is given by

  $S=N-P.$

  $\Rightarrow S=\left(1.25t^2-t+7.5\right)-\left(0.7t^2-0.95t+3.8\right)$

  $\Rightarrow S=1.25t^2-t+7.5-0.7t^2+0.95t-3.8$

  $\Rightarrow S=1.25t^2-0.7t^2-t+0.95t+7.5-3.8$

  $\Rightarrow S=0.55t^2-0.05t+3.7$

Conclusion:

Therefore, number of studentsin grades 6 through 12 is modeled by the equation

  $S=0.55t^2-0.05t+3.7$ .

b.

To determine

To find the number of students enrolled in grades 6 through 12 in the year 2013.

Answer to Problem 51PPS

  $30$

Explanation of Solution

Given:

Total number of students in school enrolled from 2006 to 2015, $N=1.25t^2-t+7.5$ .

Number of students in Kindergarten through fifth grade, $P=0.7t^2-0.95t+3.8$ .

Number of students in grades 6 through 12 from 2006 to 2015, $S=0.55t^2-0.05t+3.7$ .

Concept used:

Number of students in grades 6 through 12 from 2006 to 2015, $S=0.55t^2-0.05t+3.7$ .

Calculation: Number of students in grades 6 through 12 from 2006 to 2015, $S=0.55t^2-0.05t+3.7$ .

Number of years, $t=2013-2006$

         $=7$

  $\begin{array}{l}\Rightarrow S=0.55t^2-0.05t+3.7\\ \Rightarrow S=0.55\times {\left(7\right)}^2-0.05\times \left(7\right)+3.7\\ \Rightarrow S=0.55\times 49-0.05\times 7+3.7\\ \Rightarrow S=26.95-0.35+3.7\\ \Rightarrow S=26.95+3.7-0.35\\ \Rightarrow S=30.65-0.35\\ \Rightarrow S\approx 30\end{array}$

Conclusion:

Therefore, the number of students in the year 2013 is $30$ .