1. **Problem ①:** A farm stand sells oranges in 3-lb bags for 3.75. We want to find an equation for cost $c$ in terms of pounds $p$. Since a 3-lb bag costs 3.75, cost per pound is $$\frac{3.75}{3} = 1.25.$$ So the cost $c$ for $p$ pounds is $$c = 1.25p.$$ This matches choice A. 2. **Problem ②:** Are the photo dimensions proportional? We check ratios of width to height for each photo: - Student Photo: width = 1, height = $1\frac{1}{2} = 1.5$, ratio = $\frac{1}{1.5} = \frac{2}{3}$. - Small Photo: width = 2, height = $2\frac{1}{2} = 2.5$, ratio = $\frac{2}{2.5} = \frac{4}{5}$. - Medium Photo: width = 4, height = 5, ratio = $\frac{4}{5}$. - Large Photo: width = 8, height = 10, ratio = $\frac{8}{10} = \frac{4}{5}$. Since the student photo ratio $\frac{2}{3}$ differs from $\frac{4}{5}$ for the other photos, the dimensions are not all proportional. Jessica is not correct. 3. **Problem ③:** Victor runs 3 laps every 5 minutes. Fill the table: Given laps: blank and 1 1/5 (which is 1.2 laps), and times: 1 and 6 minutes. First, find time for 1 lap: $$\text{Rate} = \frac{3\text{ laps}}{5\text{ min}} = 0.6\text{ laps per minute}.$$ Time for 1 lap is $$\frac{1}{0.6} = \frac{5}{3} = 1\frac{2}{3} \text{ minutes} \approx 1.67.$$ Time for 1.2 laps is $$1.2 \times \frac{5}{3} = 2 \text{ minutes}.$$ Filling table: - Laps: 0.6 laps (to correspond to time 1 min), and 1.2 laps (given). - Times: 1 min (given), and 6 min (given). Since 3 laps take 5 minutes, for 6 minutes: $$\text{laps} = 0.6 \times 6 = 3.6 \text{ laps}.$$ Thus, the corrected table is: | Laps | 0.6 | 1.2 | 3.6 | | Time (min) | 1 | 2 | 6 | But the user table has only two columns: blank and 1 1/5 laps with times 1 and 6. Let’s fill blank lap for 1 min: $$\text{Laps} = 0.6 \times 1 = 0.6.$$ Final table: | Laps | 0.6 | 1.2 | | Time (min) | 1 | 6 | **Summary:** - Problem ① answer: $c = 1.25p$ - Problem ②: Dimensions are NOT proportional because ratios differ. - Problem ③: For 1 min, laps = 0.6. For 6 min, laps = 3.6.