Answered

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A certain bacteria grows 75% every hour. A sick child starts with 10 bacteria cells. How many cells will be in the child’s body at the end of one day? A. 75 B. 75000C. 6,806,333D. -75000

Sagot :

This is a geometric sequence problem.

The first term is 10 bacteria cells

and it grows 75% every hour.

It can be represented as :

[tex]n=n_1(1+r)^t[/tex]

where n is the number of cells after t hours

n1 is the first term

and

r is the growth rate.

Substitute the following values :

n1 = 10

r = 75% or 0.75

t = 1 day which is equal to 24 hrs

The number of cells after 24 hours will be :

[tex]\begin{gathered} n=n_1(1+r)^t_{} \\ n=10(1+0.75)^{24} \\ n=10(1.75)^{24} \\ n=6,806,332.61 \end{gathered}[/tex]

Therefore, the correct answer is Choice C. 6,806,333

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