Question

the number of bacteria in a culture is given by the function ( n(t)=925 e^{0.15 t} ) 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

The general form of an exponential growth function is \(n(t)=n_0e^{rt}\), where \(r\) is the relative rate of growth.
Comparing \(n(t) = 925e^{0.15t}\) with \(n(t)=n_0e^{rt}\), we get \(r = 0.15\).
To convert to a percentage, we multiply by \(100\): \(0.15\times100=15\%\)

Step2: Find the initial population

When \(t = 0\), substitute into \(n(t)=925e^{0.15t}\)
\(n(0)=925e^{0.15\times0}\)
Since \(e^{0}=1\), then \(n(0)=925\times1 = 925\)

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

Substitute \(t = 5\) into \(n(t)=925e^{0.15t}\)
\(n(5)=925e^{0.15\times5}=925e^{0.75}\)
Using a calculator, \(e^{0.75}\approx2.117\)
\(n(5)=925\times2.117 = 1958.225\approx1958\)

Answer:

(a) \(15\)
(b) \(925\)
(c) \(1958\)