Question

Can someone help me pleasee

The number of bacteria in a culture is given by the function
n(t)=990e^0.35t
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

Answer:(a) The relative rate of growth is given by the derivative of the population function:

Step-by-step explanation:n'(t) = 0.35 * 990e^0.35t

To find the relative rate of growth, we divide the derivative by the population and multiply by 100%:

n'(t)/n(t) * 100% = 0.35 * 990e^0.35t / (990e^0.35t) * 100%

= 0.35 * 100%

= 35%

Therefore, the relative rate of growth of the bacterium population is 35%.

(b) The initial population of the culture is given by the population function at t = 0:

n(0) = 990e^0.35(0)

= 990 * e^0

= 990 * 1

= 990

Therefore, the initial population of the culture is 990 bacteria.

(c) To find the number of bacteria at time t = 5, we plug in t = 5 into the population function:

n(5) = 990e^0.35(5)

= 990 * e^1.75

≈ 132426.28

Therefore, the culture will contain approximately 132426 bacteria at time t = 5 hours.