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p(t)=100(12)t8.02p(t)=100\left(\dfrac{1}{2}\right)^{\dfrac{t}{8.02}} \newlineRadioactive isotopes are unstable atoms that decay into other atoms. Iodine−131-131 is a radioactive isotope. The function models pp, the percentage of iodine−131-131 atoms remaining in a sample after tt days. To the nearest tenth of a percent, what percent of iodine−131-131 atoms remain in the sample after 1010 days?

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Q. p(t)=100(12)t8.02p(t)=100\left(\dfrac{1}{2}\right)^{\dfrac{t}{8.02}} \newlineRadioactive isotopes are unstable atoms that decay into other atoms. Iodine−131-131 is a radioactive isotope. The function models pp, the percentage of iodine−131-131 atoms remaining in a sample after tt days. To the nearest tenth of a percent, what percent of iodine−131-131 atoms remain in the sample after 1010 days?
  1. Identify Function and Value: Identify the given function and the value to be substituted.\newlineThe function given is p(t)=100×(12)t8.02p(t) = 100 \times \left(\frac{1}{2}\right)^{\frac{t}{8.02}}, which models the percentage of iodine−131-131 atoms remaining after tt days. We need to find the percentage remaining after 1010 days, so we will substitute tt with 1010.
  2. Substitute Value: Substitute the value of tt into the function.p(10)=100×(12)108.02p(10) = 100 \times \left(\frac{1}{2}\right)^{\frac{10}{8.02}}
  3. Calculate Exponent: Calculate the exponent part of the function.\newline(108.02)=1.247(\frac{10}{8.02}) = 1.247\newline(12)1.247(\frac{1}{2})^{1.247}
  4. Evaluate Exponent: Evaluate the exponent. (12)1.247≈0.435(\frac{1}{2})^{1.247} \approx 0.435
  5. Multiply by 100100: Multiply the result by 100100 to find the percentage. p(10)=100×0.435p(10) = 100 \times 0.435
  6. Perform Multiplication: Perform the multiplication to get the final answer. p(10)≈43.5%p(10) \approx 43.5\%

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