the number of bacteria in a culture is given by the function $n(t)=945e^{0.35t}$ where $t$ is measured in…

the number of bacteria in a culture is given by the function $n(t)=945e^{0.35t}$ 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

the number of bacteria in a culture is given by the function $n(t)=945e^{0.35t}$ 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

Explanation:

Step1: Find the relative rate of growth (a)

The general form of an exponential growth function is (n(t)=n_0e^{rt}), where (r) is the relative rate of growth. Comparing (n(t) = 945e^{0.35t}) with (n(t)=n_0e^{rt}), we get (r = 0.35). To convert to a percentage, we multiply by (100), so (0.35\times100=35%).

Step2: Find the initial population (b)

When (t = 0), substitute into (n(t)=945e^{0.35t}). Using the property (e^{0}=1), we have (n(0)=945e^{0}=945\times1 = 945).

Step3: Find the population at (t = 5) (c)

Substitute (t = 5) into (n(t)=945e^{0.35t}). (n(5)=945e^{0.35\times5}=945e^{1.75}). Using a calculator, (e^{1.75}\approx5.7546). Then (n(5)=945\times5.7546\approx5448).

Answer:

a. (35) b. (945) c. (5448)