homework 6.1 exponential functions\nscore: 10/140 answered: 1/14\nquestion 2\nthe number of bacteria in a…

homework 6.1 exponential functions\nscore: 10/140 answered: 1/14\nquestion 2\nthe number of bacteria in a culture is given by the function\n$n(t)=945e^{0.35t}$\nwhere $t$ is measured in hours.\n(a) what is the relative rate of growth of this bacterium population?\nyour answer is\\(\\square\\) percent\n(b) what is the initial population of the culture (at $t = 0$)?\nyour answer is\\(\\square\\)\n(c) how many bacteria will the culture contain at time $t = 5$?\nyour answer is\\(\\square\\)

homework 6.1 exponential functions\nscore: 10/140 answered: 1/14\nquestion 2\nthe number of bacteria in a culture is given by the function\n$n(t)=945e^{0.35t}$\nwhere $t$ is measured in hours.\n(a) what is the relative rate of growth of this bacterium population?\nyour answer is\\(\\square\\) percent\n(b) what is the initial population of the culture (at $t = 0$)?\nyour answer is\\(\\square\\)\n(c) how many bacteria will the culture contain at time $t = 5$?\nyour answer is\\(\\square\\)

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 the relative rate of growth is (0.35\times100=35%).

Step2: Find the initial population (b)

For the initial population, we substitute (t = 0) into the function (n(t)). When (t=0), (n(0)=945e^{0.35\times0}). Since (e^{0}=1), then (n(0)=945\times1 = 945).

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

Substitute (t = 5) into the function (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) percent (b) (945) (c) (5448)