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In a lab experiment, the decay of a radioactive isotope is being observed. At the beginning of the first day of the experiment the mass of the substance was 1000 grams and mass was decreasing by 9% per day. Determine the mass of the radioactive sample at the beginning of the 12th day of the experiment. Round to the nearest tenth (if necessary). Answer: □ grams

In a lab experiment, the decay of a radioactive isotope is being observed. At the beginning of the first day of the experiment the mass of the substance was 10001000 grams and mass was decreasing by 9% 9 \% per day. Determine the mass of the radioactive sample at the beginning of the 1212th day of the experiment. Round to the nearest tenth (if necessary).\newlineAnswer: □\square grams

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Q. In a lab experiment, the decay of a radioactive isotope is being observed. At the beginning of the first day of the experiment the mass of the substance was 10001000 grams and mass was decreasing by 9% 9 \% per day. Determine the mass of the radioactive sample at the beginning of the 1212th day of the experiment. Round to the nearest tenth (if necessary).\newlineAnswer: □\square grams
  1. Identify initial mass and decrease: Identify the initial mass and the daily percentage decrease. The initial mass of the radioactive substance is 10001000 grams, and it decreases by 9%9\% each day.
  2. Determine decay factor: Determine the decay factor.\newlineSince the mass decreases by 9%9\% each day, the decay factor is 1−0.09=0.911 - 0.09 = 0.91.
  3. Calculate mass at 1212th day: Calculate the mass at the beginning of the 12th12^{\text{th}} day.\newlineWe use the formula for exponential decay: P(t)=P0×(decay factor)tP(t) = P_0 \times (\text{decay factor})^{t}, where P(t)P(t) is the mass at time tt, P0P_0 is the initial mass, and tt is the number of days.
  4. Substitute values into formula: Substitute the values into the formula. P(11)=1000×0.9111P(11) = 1000 \times 0.91^{11}, because we are looking for the mass at the beginning of the 1212th day, which is after 1111 complete days.
  5. Perform calculation: Perform the calculation.\newlineP(11)=1000×0.9111P(11) = 1000 \times 0.91^{11}\newlineP(11)≈1000×0.31381059609P(11) \approx 1000 \times 0.31381059609 (using a calculator)\newlineP(11)≈313.8P(11) \approx 313.8 grams
  6. Round to nearest tenth: Round to the nearest tenth.\newlineThe mass of the radioactive sample at the beginning of the 12th12^{\text{th}} day is approximately 313.8313.8 grams when rounded to the nearest tenth.

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