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.​

Answer 1

Answer :

Given:

Intial count of Bacteria (P) = 5,06,000, increasing Rate of Bacteria(R) = 2.5%, Time = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P(1+R/100)^n

A= 506000(1+2.5/100)^n

A= 506000(1+25/1000)²

A= 506000(1+1/40)²

A= 506000(41/40)²

A= 506000(41×41/40×40)

A= 506000×1681/ 1600

A=(5060 × 1681) /16

A= 316.25×1681

A= 5,31,616.25

Hence, the number of bacteria after two hours are 531616 (approx.).

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Answer 2

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 ?

Answer :—

P ↦ 5,06,000

R ↦ 2.5%

T ↦ 2hA

= p($1 \times \frac{2.5}{100}$)²h

» 5,06,000 × 41/40×41/40

» 8505860/10

» 531616 (Approx).