The Count of bacteria in a certain experiment increased at the rate of 6% per hour find the count of bacteria at the end of 3 hours if it was 4500000 in the beginning
Answers
Answered by
1
Answer:
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 (approx)
Answered by
11
Answer:
5359572.0
Step-by-step explanation
PRINCIPAL=4500000
RATE=6%
TIME=3HOURS
A=PRINCIPAL{1+R/100}^N
A=4500000{1+6/100}^3
A=4500000{100+6/100}^3
A=4500000[106+/100]^3
A=4500000X106/100X106/100X106/100
A=45X106X106X106/10
A=53595720/10
A=5359572.0
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