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Exercise 5.2 · Q15

Q.A certain type of bacteria doubles its population every 20 minutes. Assuming no bacteria die, how many bacteria will there be after 3 hours if there are 1 million bacteria at present?

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Each 20-minute period doubles the population; find how many 20-minute periods fit in 3 hours and apply N=N0⋅2nN=N_0\cdot2^n.

N=N0⋅2nN=N_0\cdot2^{n}, where N0N_0 = initial population, nn = number of doubling periods.

  1. N0=1,000,000N_0=1{,}000{,}000; doubling time =20=20 min.
  2. Total time =3=3 hours =180=180 minutes.
  3. Number of doubling periods: n=18020=9n=\dfrac{180}{20}=9. …

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