1. **State the problem:** We want to approximate the average amount of water in the tank over the interval $[0,12]$ using a right Riemann sum with 4 subintervals. 2. **Given data:** - Time values: $t = 0, 3, 7, 9, 12$ - Corresponding amounts: $A(t) = 36, 51, 43, 49, 64$ 3. **Right Riemann sum formula:** $$\text{Right Riemann sum} = \sum_{i=1}^n A(t_i) \Delta t_i$$ where $\Delta t_i$ is the width of each subinterval and $A(t_i)$ is the function value at the right endpoint of each subinterval. 4. **Calculate subinterval widths:** - $\Delta t_1 = 3 - 0 = 3$ - $\Delta t_2 = 7 - 3 = 4$ - $\Delta t_3 = 9 - 7 = 2$ - $\Delta t_4 = 12 - 9 = 3$ 5. **Apply right endpoints:** The right endpoints are $t=3,7,9,12$ with values $51,43,49,64$ respectively. 6. **Calculate the right Riemann sum:** $$ \begin{aligned} \text{Sum} &= 51 \times 3 + 43 \times 4 + 49 \times 2 + 64 \times 3 \\ &= 153 + 172 + 98 + 192 \\ &= 615 \end{aligned} $$ 7. **Calculate the average amount of water:** Average amount = $\frac{\text{Sum}}{12 - 0} = \frac{615}{12} = 51.25$ **Final answer:** The average amount of water in the tank over $[0,12]$ is approximately **51.25 liters**.