1. **Problem statement:** The amount of a medicinal drug in the body $t$ hours after injection is given by $$D(t) = 23(0.85)^t, \quad t \geq 0,$$ with zero amount before injection. 2. **(a.i) Initial dose:** The initial dose is the amount at $t=0$. Substitute $t=0$ into the formula: $$D(0) = 23(0.85)^0 = 23 \times 1 = 23$$ milligrams. 3. **(a.ii) Percentage of drug leaving each hour:** The drug amount decreases by a factor of $0.85$ each hour, meaning $85\%$ remains. The percentage leaving is: $$100\% - 85\% = 15\%$$ per hour. 4. **(b) Amount remaining after 10 hours:** Substitute $t=10$: $$D(10) = 23(0.85)^{10}.$$ Calculate: $$0.85^{10} \approx 0.1969,$$ so $$D(10) \approx 23 \times 0.1969 = 4.53$$ milligrams (rounded to two decimal places). **Final answers:** - Initial dose: $23$ mg - Percentage leaving per hour: $15\%$ - Amount after 10 hours: approximately $4.53$ mg