1. **State the problem:** We are given the function $$p(t) = 100 \left(\frac{1}{2}\right)^{\frac{t}{8.02}}$$ which models the percentage of iodine-131 atoms remaining after $$t$$ days. We need to find the percentage remaining after 10 days. 2. **Formula and explanation:** This is an exponential decay function where the base $$\frac{1}{2}$$ represents the half-life decay factor, and 8.02 is the half-life period in days. 3. **Substitute $$t=10$$ into the function:** $$p(10) = 100 \left(\frac{1}{2}\right)^{\frac{10}{8.02}}$$ 4. **Calculate the exponent:** $$\frac{10}{8.02} \approx 1.247$$ 5. **Evaluate the power:** $$\left(\frac{1}{2}\right)^{1.247} = 2^{-1.247}$$ 6. **Calculate the value:** $$2^{-1.247} = \frac{1}{2^{1.247}} \approx \frac{1}{2.37} \approx 0.4219$$ 7. **Calculate the percentage remaining:** $$p(10) = 100 \times 0.4219 = 42.19$$ 8. **Round to the nearest tenth:** $$42.2$$ **Final answer:** Approximately 42.2 percent of iodine-131 atoms remain after 10 days.