Answer:
(a) The equation for the number of bacteria at time t ≥ 0 is 
(b) The average population in the bacteria for 0 ≤ t ≤ 10 is 29,482 bacteria
(c) The growth rate of the bacteria in the first 10 hours is 14,841 Bacteria/hour
Step-by-step explanation:
(a) The given rate of growth of the bacteria population is;

The initial amount of bacteria = 200
At time, t ≥ 0, we have;


ln(y) = 0.5·t + C


When, t = 0, y = 200, we have;

C₁ = 200
The equation for the number of bacteria at time t ≥ 0 is therefore given as follows;

(b) The expression for the average number of bacteria in the population for 0 ≤ t ≤ 10 is given as follows;

![y = \int\limits^{10}_0 {0.5 \cdot 200\cdot e^{0.5 \cdot t }} \, dt = \left[200\cdot e^{0.5 \cdot t }\right]^{10}_0 \approx 2,9682.6 - 200 = 29,482.6](https://tex.z-dn.net/?f=y%20%3D%20%5Cint%5Climits%5E%7B10%7D_0%20%7B0.5%20%5Ccdot%20%20200%5Ccdot%20e%5E%7B0.5%20%5Ccdot%20t%20%7D%7D%20%5C%2C%20dt%20%3D%20%5Cleft%5B200%5Ccdot%20e%5E%7B0.5%20%5Ccdot%20t%20%7D%5Cright%5D%5E%7B10%7D_0%20%5Capprox%202%2C9682.6%20-%20200%20%3D%2029%2C482.6)
The average population in the bacteria for 0 ≤ t ≤ 10 is therefore, y = 29,482
(c) The average rate of growth after 10 hours of growth is given as follows;


The growth rate of the bacteria in the first 10 hours is 14,841 Bacteria/hour