Question

The number of bacteria in a culture is given by the function: n(t)=905e^.35tn(t)=905e0.35t 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

Answer 1

Check below, please.

1) Since the function is modeled like this:

`N(t)=905e^(0.35t)`

a) Let's plug into that the given data and find the relative growth

`\begin{gathered} N(1)=1284.2 \\ N(2)=1822.4 \\ (1822.4)/(1284.2)=1.4190-1=0.4190\Rightarrow41.9\% \end{gathered}`

Plugging into that t=1 and t=2 we could find two values, and dividing we found that there was a growth of approximately 41.9% in the Bacterium population.

b) The initial population of the culture is found when we plug t=0

`N(0)=905\left(e\right)^(0.35t)\Rightarrow N(0)=905`

c) Let's plug into t=5 and find the number:

`\begin{gathered} N(5)=905e^(0.35(5)) \\ N(5)=5207.91542\approx5208 \end{gathered}`