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Initially, there are 4×10^(3) bacteria in a container. It is given that the number of bacteria will be doubled after every 20 minutes. If the container is left alone for 6 hours, find the number of bacteria in the container, (Express the answer in scientific notation correct to 3 significant figures.)

Initially, there are 4×103 4 \times 10^{3} bacteria in a container. It is given that the number of bacteria will be doubled after every 2020 minutes. If the container is left alone for 66 hours, find the number of bacteria in the container, (Express the answer in scientific notation correct to 33 significant figures.)

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Full solution

Q. Initially, there are 4×103 4 \times 10^{3} bacteria in a container. It is given that the number of bacteria will be doubled after every 2020 minutes. If the container is left alone for 66 hours, find the number of bacteria in the container, (Express the answer in scientific notation correct to 33 significant figures.)
  1. Calculate Doubling Periods: Calculate the total number of doubling periods in 66 hours. \newline66 hours =360= 360 minutes. \newlineDoubling period =20= 20 minutes. \newlineNumber of periods =360/20=18= 360 / 20 = 18.
  2. Apply Doubling Formula: Apply the doubling formula to find the final number of bacteria.\newlineInitial count = 4×1034\times10^3. \newlineFinal count = Initial count ×2Number of periods\times 2^{\text{Number of periods}}.\newlineFinal count = 4×103×2184\times10^3 \times 2^{18}.
  3. Calculate Final Count: Calculate 2182^{18} and multiply by the initial count.\newline218=2621442^{18} = 262144.\newlineFinal count = 4×103×262144=1048576×1034\times10^3 \times 262144 = 1048576\times10^3.

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