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

Based on the model, percentage of increasing of bacteria in the pond per year is given by, 59% .

Explanation:

According to the question,

A(t) =
`136 * (1.123)^(4t)` ---------------------(1)

where, A(t) = number of E-coli bacteria cells in the pond per 100 mL of water at time t year.

So,

A(t+1) =
`136 * (1.123)^((4(t+1))`

`\simeq 136 * 1.59 * (1.123)^((4t))`--------------------(2)

So, based on the model, percentage of increasing of bacteria in the pond per year is given by,

`(\frac {(A(t+1) - A(t))}{A(t)} * 100)` %

`\simeq 59` % [from (1) and (2)]