Question

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 45.00.000 in the beginning​

Answer 1

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$\sf \: P_1$

= Original amount of bacteria= 4500000

Rate of increase= 6% per hour

Time =3 hrs

Let Bacteria count after three hours be the P

$\sf \: P = P_1(1 + \frac{R}{100} {)}^{t}$

$\sf \: 4500000 = P(1 + \frac{6}{100} {)}^{3}$

$\sf \: P = 4500000 \times \frac{6}{100} \times \frac{6}{100} \times \frac{6}{100} = 45349632$

So after 3 hours the count of bacteria will be 45349632

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