Answer:
P = 5, 06, 000 R = 2.5% per hour \[\therefore \] \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=5,06,000{{\left( 1+\frac{2.5}{100} \right)}^{2}}\] \[=5,06,000{{\left( 1+\frac{1}{40} \right)}^{2}}\] \[=5,06,000{{\left( \frac{41}{40} \right)}^{2}}\] \[=5,06,000\times \frac{41}{40}\times \frac{41}{40}\] = 5,31, 616 (approx.) Hence, the count of bacteria at the end of 2 hours is 5, 31, 616 (approx.).
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