Question

He 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.

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

A.
59%
B.
40%
C.
41%
D.
60%

Answer 1

The function that represents the number of E.coli bacteria cells per 100 mL of water as t years elapses, and is missing in the question, is:

`A(t)=136(1.123)^(4t)`

Answer:

Option A. 59%

Step-by-step explanation:

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::

• 
  `A(t)=136(1.123)^(4t)`

• 
  `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.

You can calculate it also using two consecutive values for t. For instance, use t =1 and t = 1

`t=1\\\\ A(1)=136(1.123)^4\\\\\\t=2\\ \\ \\A(2)=136(1.123)^8\\ \\ \\A(2)/A(1)=1.123^8/1.1123^4=1.123^4=1.590`