Ountrinic Epronamial und Lepricimic forctions Evelhating an exponential function that models a feul worid sitimition The radioactive substance cesium-137 has a half-life of 30 years. The amount \( A \) ( \( t \) ) (in grams) of a sample of cesium-137 remaining after tyears is given by the following exponential function. \[ A(t)=621\left(\frac{1}{2}\right)^{\frac{t}{30}} \] Find the initial amount in the sample and the amount remaining after 100 years. Round your answers to the nearest gram as necessary. Initial amount: grams Amount after 100 years: \( \square \) grams

To solve the problem of determining the initial amount of cesium-137 in the sample and the amount remaining after 100 years, we'll use the given exponential decay function: \[ A(t) = 621 \left(\frac{1}{2}\right)^{\frac{t}{30}} \] where: - \( A(t) \) is the amount of cesium-137 remaining after \( t \) years. - \( t \) is the time in years. - 621 grams is the initial amount of cesium-137. ### 1. **Initial Amount in the Sample** The initial amount is the amount of cesium-137 present at time \( t = 0 \). \[ A(0) = 621 \left(\frac{1}{2}\right)^{\frac{0}{30}} = 621 \times 1 = 621 \text{ grams} \] **Initial amount:** **621 grams** ### 2. **Amount Remaining After 100 Years** To find the amount remaining after 100 years, substitute \( t = 100 \) into the function: \[ A(100) = 621 \left(\frac{1}{2}\right)^{\frac{100}{30}} = 621 \left(\frac{1}{2}\right)^{3.\overline{3}} \] First, simplify the exponent: \[ \frac{100}{30} = \frac{10}{3} \approx 3.3333 \] Now, calculate \( \left(\frac{1}{2}\right)^{3.3333} \): \[ \left(\frac{1}{2}\right)^{3.3333} = 2^{-3.3333} \approx 0.0992 \] Multiply this by the initial amount: \[ A(100) \approx 621 \times 0.0992 \approx 61.6 \text{ grams} \] Rounding to the nearest gram: **Amount after 100 years:** **62 grams** ### **Summary** - **Initial amount:** 621 grams - **Amount after 100 years:** 62 grams