Question

the number of bacteria in a culture is given by the function (n(t)=980e^{0.15t}) 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 question help: message instructor post to forum

Explanation:

Step1: Identify the growth - rate formula

The general form of an exponential growth function is $n(t)=n_0e^{rt}$, where $r$ is the relative rate of growth. In the given function $n(t) = 980e^{0.15t}$, the value of $r$ is $0.15$.

Step2: Convert $r$ to percentage for part (a)

To convert a decimal to a percentage, we multiply by 100. So, $0.15\times100 = 15$.

Step3: Find the initial population for part (b)

Substitute $t = 0$ into the function $n(t)=980e^{0.15t}$. Since $e^{0}=1$, then $n(0)=980e^{0.15\times0}=980\times1 = 980$.

Step4: Find the population at $t = 5$ for part (c)

Substitute $t = 5$ into the function $n(t)=980e^{0.15t}$. So, $n(5)=980e^{0.15\times5}=980e^{0.75}$. Using a calculator, $e^{0.75}\approx2.117$, and $n(5)=980\times2.117 = 2074.66\approx2075$.

Answer:

(a) 15 percent
(b) 980
(c) 2075