Chapter 11, Problem 78SGA

a.

To determine

To find: The average number of words per minute after 2 weeks.

Answer to Problem 78SGA

The average number of words per minute after 2 weeks is $45$ .

Explanation of Solution

Given information: The average number of words per minute, $N$ , input by trainees after $t$ weeks $N=65-30e^{-0.20t}$ .

Calculation:

Consider the equation.

  $\begin{array}{c}N=65-30e^{-0.20t}\\ N=65-30e^{-0.20\left(2\right)}\\ =65-20\\ =45\end{array}$

Thus, the average number of words per minute after 2 weeks is $45$ .

b.

To determine

To find: The average number of words per minute after 15 weeks.

Answer to Problem 78SGA

The average number of words per minute after 15 weeks is $64$ .

Explanation of Solution

Given information: The average number of words per minute, $N$ , input by trainees after $t$ weeks $N=65-30e^{-0.20t}$ .

Calculation:

Consider the equation.

  $\begin{array}{c}N=65-30e^{-0.20t}\\ N=65-30e^{-0.20\left(15\right)}\\ =65-1\\ =64\end{array}$

Thus, the average number of words per minute after 15 weeks is $64$ .

c.

To determine

To find: The time when was average traineee able to input 50 words per minute.

Answer to Problem 78SGA

The time when was average traineee able to input 50 words per minute is $3.5\text{ weeks}$ .

Explanation of Solution

Given information: The average number of words per minute, $N$ , input by trainees after $t$ weeks $N=65-30e^{-0.20t}$ .

Calculation:

Consider the equation.

  $\begin{array}{c}N=65-30e^{-0.20t}\\ 30e^{-0.20t}=65-N\\ e^{-0.20t}=\frac{65-N}{30}\\ \ln e^{-0.20t}=\ln \frac{65-N}{30}\\ -0.2t=\ln \frac{65-N}{30}\\ t=-\frac{\ln \frac{65-N}{30}}{0.2}\end{array}$

Substitute the value in above equation.

  $\begin{array}{c}t=-\frac{\ln \frac{65-50}{30}}{0.2}\\ =-\frac{\ln \frac{1}{2}}{0.2}\\ =3.5\text{ weeks}\end{array}$

Thus, the time when was average traineee able to input 50 words per minute is $3.5\text{ weeks}$ .