. 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 2 hours if the count was initially
5.06.000. (Hint: Bacterial growth is an
example of compound interest)
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Answers
Answered by
1
Step-by-step explanation:
= Original count of bacteria =506000,
Rate of increase =R=2.5% per hour,
Time =2 hours
∴ Bacteria count after 2 hours = P
P=P
0
(1+
100
R
)
T
=506000×(1+
100
2.5
)
2
=506000×
100
102.5
×
100
102.5
=531616.25=531616 (approx)
Hope this helps!
Answered by
0
Answer:
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 2 hours, if the count was initially 5,06,000.
We have, P
0
= Original count of bacteria =506000,
Rate of increase =R=2.5% per hour,
Time =2 hours
∴ Bacteria count after 2 hours = P
P=P
0
(1+
100
R
)
T
=506000×(1+
100
2.5
)
2
=506000×
100
102.5
×
100
102.5
=531616.25=531616
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