In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour.Find the bacteria at the end of 1 hour of the count was initially 5,060.
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Answered by
2
Answer:
here is the explanation bro...
Step-by-step explanation:
Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours
After 2 hours, number of bacteria,
Amount (A) = P\left(1+\frac{R}{100}\right)^nP(1+
100
R
)
n
= 506000\left(1+\frac{2.5}{100}\right)^2506000(1+
100
2.5
)
2
= 506000\left(1+\frac{25}{1000}\right)^2506000(1+
1000
25
)
2
= 506000\left(1+\frac{1}{40}\right)^2506000(1+
40
1
)
2
= 506000\left(\frac{41}{40}\right)^2506000(
40
41
)
2
= 506000\times\frac{41}{40}\times\frac{41}{40}506000×
40
41
×
40
41
= 5,31,616.25
Hence, number of bacteria after two hours are 531616 (approx.).
Answered by
0
Hence, number of bacteria after two hours are 531616 (approx.).
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