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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 1400 grams and mass was decreasing by 12% per day. Determine the mass of the radioactive sample at the beginning of the 18th 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 14001400 grams and mass was decreasing by 12% 12 \% per day. Determine the mass of the radioactive sample at the beginning of the 1818th 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 14001400 grams and mass was decreasing by 12% 12 \% per day. Determine the mass of the radioactive sample at the beginning of the 1818th 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.\newlineThe initial mass of the radioactive substance is 14001400 grams, and it decreases by 12%12\% each day.
  2. Determine decay factor: Determine the decay factor.\newlineSince the mass decreases by 12%12\% each day, the decay factor is 100%−12%=88%100\% - 12\% = 88\% per day. To use this in calculations, convert the percentage to a decimal by dividing by 100100. Thus, the decay factor is 0.880.88.
  3. Calculate mass at 1818th day: Calculate the mass at the beginning of the 1818th day.\newlineThe formula for exponential decay is given by P(t)=P0×(decay factor)tP(t) = P_0 \times (\text{decay factor})^{t}, where P(t)P(t) is the final mass, P0P_0 is the initial mass, and tt is the time in days.\newlineFor this problem, P0=1400P_0 = 1400 grams, the decay factor is 0.880.88, and t=17t = 17 (since we want the mass at the beginning of the 1818th day, we use t=17t = 17 days).\newlineP(17)=1400×(0.88)17P(17) = 1400 \times (0.88)^{17}
  4. Perform calculation: Perform the calculation.\newlineP(17)=1400×(0.88)17P(17) = 1400 \times (0.88)^{17}\newlineUsing a calculator, we find:\newlineP(17)≈1400×0.114P(17) \approx 1400 \times 0.114\newlineP(17)≈159.6P(17) \approx 159.6
  5. Round result: Round the result to the nearest tenth. The mass of the radioactive sample at the beginning of the 18th18^{\text{th}} day is approximately 159.6159.6 grams when rounded to the nearest tenth.

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