Question

The number of E.coli bacteria cells in a pond of stagnant water can be represented by the function below, where A represents the number of E.coli bacteria cells per 100 mL of water and t represents the time, in years, that has elapsed.
A(t)=136(1.123)^4t

Based on the model, by approximately what percent does the number of E.coli bacteria cells increase each year?


A.

60%

B.

59%

C.

41%

D.

40%

Answer 1

Answer:

• Option B. 59%

Explanation:

The function that represents the number of E.coli bacteria cells per 100 mL of water as the time t years elapses is:

• $A(t)=136(1.123)^{4t}$

The base, 1.123, represents the multiplicative constant rate of change of the function, so you just must substitute 1 for t in the power part of the function:

• $rate=(1.123)^{4t}=(1.123)^4=1.590$

Then, the multiplicative rate of change is 1.590, which means that every year the number of E.coli bacteria cells per 100 mL of water increases by a factor of 1.590, and that is 1.59 - 1 = 0.590 or 59% increase.