Chapter 11.3, Problem 10MCQ

To determine

To find: The population after 2, 15 and 60 years.

Answer to Problem 10MCQ

The population after 2 years is $18.2954$ , after 15 years is $181.0036$ and after 60 year is $200$ .

Explanation of Solution

Given:

The expression for the population is $n=\frac{M}{1-b{e}^{-ct}}$ .

The value of $M=200$ , $b=20$ and $c=035$ .

Calculation:

Consider the pollution after 2 years is obtained as,

  $\begin{array}{c}n=\frac{200}{1-20e^{-\left(0.35\right)2}}\\ =\frac{200}{10.9317}\\ =18.2954\end{array}$

Consider the pollution after 15 years is obtained as,

  $\begin{array}{c}n=\frac{200}{1-20e^{-\left(0.35\right)\left(15\right)}}\\ =\frac{200}{1.1049}\\ =181.0036\end{array}$

Consider the pollution after 60 years is obtained as,

  $\begin{array}{c}n=\frac{200}{1-20e^{-\left(0.35\right)\left(60\right)}}\\ =\frac{200}{1}\\ =200\end{array}$

Thus, the population after 2 years is $18.2954$ , after 15 years is $181.0036$ and after 60 year is $200$ .