1. **State the problem:** We are given the amount of a medicinal drug in the body at time $t$ hours after injection as $$D(t) = 23(0.85)^t, \quad t \geq 0.$$ We need to find: (a.i) The initial dose of the drug. (a.ii) The percentage of the drug that leaves the body each hour. (b) The amount of the drug remaining 10 hours after injection. 2. **Formula and explanation:** The function $D(t) = 23(0.85)^t$ is an exponential decay model where: - The initial amount is the value at $t=0$. - The base $0.85$ represents the fraction of the drug remaining after each hour. - The percentage leaving each hour is $100\% - 85\% = 15\%$. 3. **Step-by-step solution:** **(a.i) Initial dose:** Evaluate $D(0)$: $$D(0) = 23(0.85)^0 = 23 \times 1 = 23$$ So, the initial dose is 23 milligrams. **(a.ii) Percentage of drug leaving each hour:** Since $0.85$ is the fraction remaining after 1 hour, the fraction leaving is: $$1 - 0.85 = 0.15$$ Convert to percentage: $$0.15 \times 100 = 15\%$$ So, 15% of the drug leaves the body each hour. **(b) Amount remaining after 10 hours:** Calculate $D(10)$: $$D(10) = 23(0.85)^{10}$$ Calculate $(0.85)^{10}$: $$0.85^{10} \approx 0.1969$$ Multiply: $$23 \times 0.1969 \approx 4.53$$ So, approximately 4.53 milligrams of the drug remain after 10 hours. **Final answers:** - Initial dose: 23 mg - Percentage leaving each hour: 15% - Amount after 10 hours: approximately 4.53 mg