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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?

Sagot :

Answer:

a. Relative rate of growth = 0.2

b. Initial population: 920

c. 2501

Explanation:

If we have an exponential function of the form

[tex]y=A_0e^{kt}[/tex]

Then

A0 = inital amount

k = relative rate of growth

t = time

Now in our case we have

[tex]n(t)=920e^{0.2t}[/tex]

Therefore,

Inital population = 920

Relative rate of growth = 0.2

Now at t = 5, the above formula gives

[tex]n(5)=920e^{0.2*5}[/tex]

which evaluates to give

[tex]n(5)=2500.8[/tex]

which rounded to the nearest whole number is

[tex]\boxed{n(5)=2501.}[/tex]

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