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:

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

Step-by-step explanation:

According to the question,

A(t) = $136 \times (1.123)^{4t}$ ---------------------(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) = $136 \times (1.123)^{(4(t+1)}$

          $\simeq 136 \times 1.59 \times (1.123)^{(4t)}$--------------------(2)

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

$(\frac {(A(t+1) - A(t))}{A(t)} \times 100)$ %

$\simeq 59$ % [from  (1) and (2)]