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Sagot :
Given:
The initial population of bacteria, I = 20
Growth rate, r = 28%
Explanation:
a) To find: The time
i) Double its size
Using the formula,
[tex]F=I(1+r)^t,\text{ Where I denotes initial and F denotes Final size.}[/tex]On substitution we get,
[tex]\begin{gathered} 40=20(1+0.28)^t \\ 1.28^t=\frac{40}{20} \\ t=\log _{1.28}2 \\ t=2.81 \end{gathered}[/tex]Thus, the answer is 2.81 hours.
ii) To reach 500000 bacteria:
[tex]\begin{gathered} 500000=20(1+0.28)^t \\ 1.28^t=\frac{500000}{20} \\ t=\log _{1.28}(25000) \\ t=41.02\text{ hours} \end{gathered}[/tex]Thus, the answer is 41.02 hours.
b) To find the bacteria size:
i) In 3 hours,
[tex]\begin{gathered} F=20(1+0.28)^3 \\ =41.94304 \\ \approx42 \end{gathered}[/tex]Thus, the size of bacteria in 3 hours is 42.
i) In 3.5 days,
That is, 84 hours
[tex]\begin{gathered} F=20(1+0.28)^{84} \\ =20261306488.67 \\ \approx20261306489 \end{gathered}[/tex]Thus, the size of bacteria in 3.5 days is 20261306489.
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