Question

the number of bacteria in a culture is given by the function $n(t)=940e^{0.15t}$ where $t$ is measured in hours. (a) what is the continuous 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 round to the nearest bacteria. question help: video

Explanation:

Step1: Find continuous growth rate

The general form of exponential growth is \(n(t)=n_0e^{rt}\), where \(r\) is the continuous growth rate.
Comparing \(n(t) = 940e^{0.15t}\) with \(n(t)=n_0e^{rt}\), we get \(r = 0.15\).
To convert to percentage, multiply by \(100\): \(0.15\times100 = 15\%\)

Step2: Find initial population

For initial population, set \(t = 0\) in \(n(t)=940e^{0.15t}\)
\(n(0)=940e^{0.15\times0}=940\times1 = 940\)

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

Substitute \(t = 5\) into \(n(t)=940e^{0.15t}\)
\(n(5)=940e^{0.15\times5}=940e^{0.75}\)
Using \(e^{0.75}\approx2.117\)
\(n(5)=940\times2.117 = 1990\) (rounded to the nearest whole number)

Answer:

(a) \(15\)
(b) \(940\)
(c) \(1990\)