1. **Problem Statement:** Find the average rate of change of the sheep's mass between $t=103$ and $t=111$ days, where $M(t) = 27.6 + 0.3t - 0.001t^2$. 2. **Formula for Average Rate of Change:** The average rate of change of a function $M(t)$ between $t=a$ and $t=b$ is given by: $$\frac{M(b) - M(a)}{b - a}$$ 3. **Calculate $M(103)$ and $M(111)$:** $$M(103) = 27.6 + 0.3(103) - 0.001(103)^2 = 27.6 + 30.9 - 0.001 \times 10609 = 27.6 + 30.9 - 10.609 = 47.891$$ $$M(111) = 27.6 + 0.3(111) - 0.001(111)^2 = 27.6 + 33.3 - 0.001 \times 12321 = 27.6 + 33.3 - 12.321 = 48.579$$ 4. **Calculate the average rate of change:** $$\frac{M(111) - M(103)}{111 - 103} = \frac{48.579 - 47.891}{8} = \frac{0.688}{8} = 0.086$$ 5. **Interpretation:** The average rate of change of the sheep's mass between 103 and 111 days is approximately $0.086$ kilograms per day. **Final answer:** $$\boxed{0.086}$$ kilograms per day.