1. **State the problem:** We have a histogram showing weights of apples in grams with frequency densities for intervals of 30 grams. 2. **Calculate the number of apples in each interval:** The number of apples in an interval is frequency density \( \times \) class width. 3. **Calculate the number of apples between 90 g and 165 g:** This covers intervals 90-120 g, 120-150 g, and part of 150-180 g. - For 90-120 g: frequency density = 1.8, width = 30, so number = \(1.8 \times 30 = 54\) - For 120-150 g: frequency density = 1.2, width = 30, so number = \(1.2 \times 30 = 36\) - For 150-180 g: frequency density = 1.0, width = 30, but only from 150 to 165 g (15 g), so number = \(1.0 \times 15 = 15\) 4. **Sum these to find total apples between 90 g and 165 g:** $$54 + 36 + 15 = 105$$ 5. **Given in the problem:** 130 apples weigh between 90 g and 165 g, so the histogram estimate is slightly off, but we use 130 as the total number of apples in this range. 6. **Calculate total number of apples:** Sum the number of apples in all intervals using frequency density and width: - 0-30 g: \(0.5 \times 30 = 15\) - 30-60 g: \(1.0 \times 30 = 30\) - 60-90 g: \(0.8 \times 30 = 24\) - 90-120 g: \(1.8 \times 30 = 54\) - 120-150 g: \(1.2 \times 30 = 36\) - 150-180 g: \(1.0 \times 30 = 30\) - 180-210 g: \(0.6 \times 30 = 18\) - 210-240 g: \(0.4 \times 30 = 12\) Total apples estimate: $$15 + 30 + 24 + 54 + 36 + 30 + 18 + 12 = 219$$ 7. **Use the ratio of green to red apples:** 7:4 means total parts = 7 + 4 = 11. Number of green apples: $$\frac{7}{11} \times 219 \approx 139.36 \approx 139$$ 8. **Calculate 20% of green apples given to charity:** $$0.20 \times 139 = 27.8 \approx 28$$ **Final answer:** Approximately 28 green apples are given to the charity.