the count of bacteria in a certain experiment was increasing at the rate of 2%per hour . find the bacteria at the end of 2 hours if the count was intially 500000. find by using compound interest method.
jasmeenkaur18:
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Answered by
3
Given,
R=2%
T=2
P=500000
A=P(1+R/100)^t
=500000(1+2/100)^2
=500000(51/50)^2
=500000*51/50*51/50
=200*51*51
=520,200
R=2%
T=2
P=500000
A=P(1+R/100)^t
=500000(1+2/100)^2
=500000(51/50)^2
=500000*51/50*51/50
=200*51*51
=520,200
Answered by
3
Answer:
Hey mate here is your answer
520200
Step-by-step explanation:
Given,
R=2%
T=2
P=500000
A=P(1+R/100)^t
=500000(1+2/100)^2
=500000(51/50)^2
=500000*51/50*51/50
=200*51*51
=520,200
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