1. **State the problem:** We want to find how many containers of each type (A, B, C) are needed to hold a total of $2 \frac{3}{4}$ cups of trail mix. 2. **Convert the mixed number to an improper fraction:** $$2 \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}$$ 3. **Recall the capacities:** - Container A holds $\frac{1}{4}$ cups - Container B holds $\frac{1}{2}$ cups - Container C holds $\frac{3}{4}$ cups 4. **Calculate how many containers of each type are needed to hold $\frac{11}{4}$ cups:** - For Container A: $$\text{Number of containers} = \frac{\frac{11}{4}}{\frac{1}{4}} = \frac{11}{4} \times \frac{4}{1} = 11$$ - For Container B: $$\text{Number of containers} = \frac{\frac{11}{4}}{\frac{1}{2}} = \frac{11}{4} \times \frac{2}{1} = \frac{22}{4} = \frac{\cancel{22}}{\cancel{4}} \times \frac{1}{1} = \frac{11}{2} = 5 \frac{1}{2}$$ - For Container C: $$\text{Number of containers} = \frac{\frac{11}{4}}{\frac{3}{4}} = \frac{11}{4} \times \frac{4}{3} = \frac{11}{\cancel{4}} \times \frac{\cancel{4}}{3} = \frac{11}{3} = 3 \frac{2}{3}$$ 5. **Interpretation:** - You need 11 containers of type A. - You need $5 \frac{1}{2}$ containers of type B (which means 5 full containers and one half-full). - You need $3 \frac{2}{3}$ containers of type C (3 full containers and two-thirds of another). **Final answer:** - Container A: 11 containers - Container B: $5 \frac{1}{2}$ containers - Container C: $3 \frac{2}{3}$ containers