At any time t ≥ 0, in hours, the rate of growth of a population of bacteria is given by dy/dt = 0.5y. Initially, there are 200 bacteria. (a) Solve for y, the number of bacteria present, at any time t ≥ 0.

(b) Write and evaluate an expression to find the average number of bacteria in the population for 0 ≤ t ≤ 10.

(c) Write an expression that gives the average rate of bacteria growth over the first 10 hours of growth. Indicate units of measure.

(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;

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

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