Question

The number of bacteria in a culture is given by the function n(t)=920e^.2twhere t is measured in hours.(a) What is the relative rate of growth of this bacterium population?(b) What is the initial population of the culture (at t=0)?(c) How many bacteria will the culture contain at time t=5?

Answer 1

a. Relative rate of growth = 0.2

b. Initial population: 920

c. 2501

Step-by-step explanation:

If we have an exponential function of the form

`y=A_0e^(kt)`

Then

A0 = inital amount

k = relative rate of growth

t = time

Now in our case we have

`n(t)=920e^(0.2t)`

Therefore,

Inital population = 920

Relative rate of growth = 0.2

Now at t = 5, the above formula gives

`n(5)=920e^(0.2*5)`

which evaluates to give

`n(5)=2500.8`

which rounded to the nearest whole number is

`\boxed{n(5)=2501.}`