Chapter 7.9, Problem 5P

a.

To determine

To calculate: The number of bacteria in a culture at 2 P.M if there were 100 bacteria at noon.

Answer to Problem 5P

The number of bacteria in a culture at 2 P.M if there were 100 bacteria at noonare $5474$ .

Explanation of Solution

Given information:

The growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

Consider the provided statement that growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

To evaluate the number of bacteria in culture at 2 P.M, substitute t as 2 in the provided expression and n as 100,

  $\begin{array}{c}N=100{\left(2.72\right)}^{2\left(2\right)}\\ =5473.63\\ \approx 5474\end{array}$

Thus, the number of bacteria in a culture at 2 P.M if there were 100 bacteria at noonare $5474$ .

b.

To determine

To calculate: The number of bacteria in a culture at 10 A.M if there were 100 bacteria at noon.

Answer to Problem 5P

The number of bacteria in a culture at 10 A.M if there were 100 bacteria at noonwere $2$ .

Explanation of Solution

Given information:

The growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

Consider the provided statement that growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

To evaluate the number of bacteria in culture at 10 A.M, substitute t as $-2$ in the provided expression and n as 100,

  $\begin{array}{c}N=100{\left(2.72\right)}^{2\left(-2\right)}\\ =1.8269\\ \approx 2.00\end{array}$

Thus, the number of bacteria in a culture at 10 A.M if there were 100 bacteria at noonwere $2$ .

c.

To determine

To calculate: The number of bacteria in a culture at 6 P.M if there were 100 bacteria at noon.

Answer to Problem 5P

The number of bacteria in a culture at 6 P.M if there were 100 bacteria at noon are $16399358$ .

Explanation of Solution

Given information:

The growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

Consider the provided statement that growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

To evaluate the number of bacteria in culture at 6 P.M, substitute t as 6 in the provided expression and n as 100,

  $\begin{array}{c}N=100{\left(2.72\right)}^{2\left(6\right)}\\ =16399358.07\\ \approx 16399358\end{array}$

Thus, the number of bacteria in a culture at 6 P.M if there were 100 bacteria at noon are $16399358$ .

d.

To determine

To calculate: The number of bacteria in a culture at 6 A.M if there were 100 bacteria at noon.

Answer to Problem 5P

The number of bacteria in a culture at 6 A.M if there were 100 bacteria at noonwere $0$ .

Explanation of Solution

Given information:

The growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

Consider the provided statement that growth rate of a bacteria culture is formulated as $N=n{\left(2.72\right)}^{2t}$ where n is the original number of bacteria and N is the number after t hours. There are 100 bacteria in culture at 12 noon.

To evaluate the number of bacteria in culture at 6 A.M, substitute t as $-6$ in the provided expression and n as 100,

  $\begin{array}{c}N=100{\left(2.72\right)}^{2\left(-6\right)}\\ =6.0977\times {10}^{-4}\\ \approx 0\end{array}$

Thus, the number of bacteria in a culture at 6 A.M if there were 100 bacteria at noonwere $0$ .