Question

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
plz don't make the answer too long

Answer 1

Answer:

2530000

Step-by-step explanation:

in one hr= 2.5%

in two hrs= 2.5% x 2 which is 5

2.5 is 506000 so 5 will be 2530000

Answer 2

Answer:

531616

Step-by-step explanation:

So, here we know that initial count was 506000.

Here, we should use the concept of compound interest.

Rate R = 2.5% per hour

Time t = 2 hrs.

Let bacteria count ( final one ) be B.

B = P*(1+R/100)^T

B = 506000*(1 + 2.5 / 100 )^2

So now, 1 + 2.5 / 100 = 100/100 + 2.5/100 = 102.5/100.

B = 506000* 102.5*102.5

                       100 * 100

B  = 5316162500 / 10000

B  = 531616.25 approximately.  531616.

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