The radioactive substance uranium- 240 has a half-life of 14 hours. The amount \( A(t) \) of a sample of uranium- 240 remaining (in grams) after \( t \) hours is given by the following exponential function. \[ A(t)=4700\left(\frac{1}{2}\right)^{\frac{t}{14}} \] Find the initial amount in the sample and the amount remaining after 60 hours. Round your answers to the nearest gram as necessary.

**Step 1. Determine the initial amount:** The initial amount is given by \( A(0) \). Substitute \( t=0 \) into the formula \[ A(t) = 4700\left(\frac{1}{2}\right)^{\frac{t}{14}}. \] Thus, \[ A(0) = 4700\left(\frac{1}{2}\right)^{\frac{0}{14}} = 4700\left(\frac{1}{2}\right)^0 = 4700 \times 1 = 4700. \] So, the initial amount of uranium-240 is **4700 grams**. **Step 2. Calculate the amount remaining after 60 hours:** Substitute \( t=60 \) into the formula: \[ A(60) = 4700\left(\frac{1}{2}\right)^{\frac{60}{14}}. \] Simplify the exponent: \[ \frac{60}{14} = \frac{30}{7} \approx 4.2857. \] Thus, \[ A(60) = 4700\left(\frac{1}{2}\right)^{4.2857}. \] **Step 3. Evaluate the exponential term:** The term can be computed as: \[ \left(\frac{1}{2}\right)^{4.2857} = 2^{-4.2857}. \] One way to compute this is using logarithms: \[ 2^{-4.2857} = e^{-4.2857\ln 2}. \] Since \(\ln 2 \approx 0.6931\), \[ -4.2857\ln 2 \approx -4.2857 \times 0.6931 \approx -2.9714. \] Thus, \[ 2^{-4.2857} \approx e^{-2.9714} \approx 0.0513. \] **Step 4. Compute \( A(60) \):** Now, \[ A(60) \approx 4700 \times 0.0513 \approx 241.11. \] Rounding to the nearest gram, \[ A(60) \approx 241 \text{ grams}. \] **Final Answer:** - The initial amount is \(4700\) grams. - The amount remaining after 60 hours is approximately \(241\) grams.