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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%

Respuesta :

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) = [tex]136 \times (1.123)^{4t}[/tex] ---------------------(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) = [tex]136 \times (1.123)^{(4(t+1)}[/tex]

          [tex]\simeq 136 \times 1.59 \times (1.123)^{(4t)}[/tex]--------------------(2)

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

[tex](\frac {(A(t+1) - A(t))}{A(t)} \times 100)[/tex] %

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

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