1. **State the problem:** We want to approximate the value of $\frac{1}{12} \int_0^{12} M(t) \, dt$ using a left Riemann sum with 4 subintervals based on the given table values of $M(t)$ at $t=0,3,6,9,12$. 2. **Recall the formula for a left Riemann sum:** $$\text{Left Riemann sum} = \sum_{i=0}^{n-1} f(x_i) \Delta x$$ where $\Delta x$ is the width of each subinterval and $f(x_i)$ is the function value at the left endpoint of each subinterval. 3. **Identify subintervals and widths:** The interval $[0,12]$ is divided into 4 subintervals: - $[0,3]$, $[3,6]$, $[6,9]$, $[9,12]$ Each subinterval has width $\Delta t = 3$ seconds. 4. **Use the left endpoints and their function values:** From the table: - $M(0) = 33$ - $M(3) = 10$ - $M(6) = 9$ - $M(9) = 10$ 5. **Calculate the left Riemann sum:** $$\sum_{i=0}^3 M(t_i) \Delta t = (33 + 10 + 9 + 10) \times 3 = 62 \times 3 = 186$$ 6. **Calculate the approximation of the integral average:** $$\frac{1}{12} \int_0^{12} M(t) \, dt \approx \frac{1}{12} \times 186 = 15.5$$ **Final answer:** $$\boxed{15.5}$$