Chapter 6.2, Problem 43E

a.

To determine

To write a function that givens the number of bacteria after $1:00p.m$

Answer to Problem 43E

The function that givens the number of bacteria after $1:00p.m$

  $N\left(t\right)=30e^{0.166t}$

Explanation of Solution

Given information:

Consider the given information,

  $N\left(t\right)=30e^{0.166t}$ Where $t$ is in hours

And is the number of cells at $t=0$

  $a=30M\text{ at 1:00pm}$

As, $a=30M$ .Tuberculosis bacteria in a sample at $\text{1:00pm}$

Therefore, the function is written as $N\left(t\right)=30e^{0.166t}$

b.

To determine

To graph the function in part (a) using graphing calculator

Explanation of Solution

Given information:

Consider the given information,

  $N\left(t\right)=30e^{0.166t}$ , where $30M$ . Is the number of bacteria?

Graph:

Use graphing calculator to graph the given function,

  

The sketched graph is given for

  $N\left(t\right)=30e^{0.166t}$

c.

To determine

To describe how to find the number of cells in the sample at 3:45 pm

Answer to Problem 43E

The number of cells in the sample at $3:45\text{pm is }47.356$

Explanation of Solution

Given information:

Consider the given information,

  $t=3:45\text{ pm}$

  $N\left(t\right)=30e^{0.166t}$ , at $t=1:00\text{ pm}$

When $t=1:00\text{ pm}$

No. of cells was $N\left(t\right)=30e^{0.166t}$

Now, at $t=3:45\text{ pm}$ , $t=2.75$ hours

No. of cells will be $\begin{array}{l}N\left(2.75\right)=30e^{0.166\times 2.75}\\ N\left(2.75\right)=47.356\end{array}$

So, the number of cells in the sample at $3:45\text{pm is }47.356$