1. **State the problem:** We have a radioactive substance with a half-life of 3 hours and an initial amount of 48,320 grams. We want to find how much remains after 15 hours. 2. **Formula used:** The amount of substance remaining after time $t$ is given by the exponential decay formula: $$ A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}} $$ where: - $A_0$ is the initial amount, - $T$ is the half-life, - $t$ is the elapsed time, - $A$ is the amount remaining after time $t$. 3. **Plug in the values:** $$ A = 48320 \times \left(\frac{1}{2}\right)^{\frac{15}{3}} $$ 4. **Simplify the exponent:** $$ \frac{15}{3} = 5 $$ So, $$ A = 48320 \times \left(\frac{1}{2}\right)^5 $$ 5. **Calculate the power:** $$ \left(\frac{1}{2}\right)^5 = \frac{1}{2^5} = \frac{1}{32} $$ 6. **Calculate the remaining amount:** $$ A = 48320 \times \frac{1}{32} $$ 7. **Show cancellation:** $$ A = 48320 \times \cancel{\frac{1}{32}} $$ 8. **Perform the division:** $$ A = 1510 $$ **Final answer:** After 15 hours, 1510 grams of the radioactive lead will remain.