In a Laboratory- the count of bacteria in a certain experiment was increasing the rate of 2-5- per hour- Find the bacteria the end of 2 hours the count was initially 5-06-000-

Question

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Initial no- of bacteria -506-000Rate -2-5-Time -2 hoursNew population of bacteria -A-506-000-1-2-5100-2-506000-1-050625-531616-25Amount of bacteria after 2 hours -531-616-

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 $5, 06, 000$ .

Initial no. of bacteria $= 506, 000$
Rate $= 2.5 %$
Time $= 2$ hours
New population of bacteria $= ?$
$A = 506, 000 {(1 + \frac{2.5}{100})}^{2} = 506000 (1.050625) = 531616.25$

Amount of bacteria after $2$ hours $= 531, 616.$

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.

Initial no. of bacteria =506,000Rate =2.5%Time =2 hoursNew population of bacteria =?A=506,000(1+2.5100)2=506000(1.050625)=531616.25Amount of bacteria after 2 hours =531,616.

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 $5, 06, 000$ .

Initial no. of bacteria $= 506, 000$
Rate $= 2.5 %$
Time $= 2$ hours
New population of bacteria $= ?$
$A = 506, 000 {(1 + \frac{2.5}{100})}^{2} = 506000 (1.050625) = 531616.25$

Amount of bacteria after $2$ hours $= 531, 616.$