Question

the number of bacteria in a culture is given by the function $n(t)=945e^{0.35t}$ where $t$ is measured in hours. (a) what is the relative rate of growth of this bacterium population? your answer is percent (b) what is the initial population of the culture (at $t = 0$)? your answer is (c) how many bacteria will the culture contain at time $t = 5$? your answer is

Explanation:

Step1: Find the relative rate of growth (a)

The general form of an exponential growth function is \(n(t)=n_0e^{rt}\), where \(r\) is the relative rate of growth.
Comparing \(n(t) = 945e^{0.35t}\) with \(n(t)=n_0e^{rt}\), we get \(r = 0.35\).
To convert to a percentage, we multiply by \(100\), so \(0.35\times100=35\%\).

Step2: Find the initial population (b)

When \(t = 0\), substitute into \(n(t)=945e^{0.35t}\).
Using the property \(e^{0}=1\), we have \(n(0)=945e^{0}=945\times1 = 945\).

Step3: Find the population at \(t = 5\) (c)

Substitute \(t = 5\) into \(n(t)=945e^{0.35t}\).
\(n(5)=945e^{0.35\times5}=945e^{1.75}\).
Using a calculator, \(e^{1.75}\approx5.7546\).
Then \(n(5)=945\times5.7546\approx5448\).

Answer:

a. \(35\)
b. \(945\)
c. \(5448\)