1. **Problem statement:** We analyze a histogram of money amounts held by 136 students. 2. **Part (a): Number of students with under 10 dollars.** - The histogram bar for 0-10 dollars is about 55 students. - Closest option is 50 (B). 3. **Part (b): Percentage of students with over 20 dollars.** - Frequencies for >20 dollars are low: roughly 10 (20-30), 5 (30-40), and 2 (40-60) students. - Total over 20 dollars = 10 + 5 + 2 = 17 students. - Percentage = $$\frac{17}{136} \times 100 \approx 12.5\%$$. - Closest option is about 10% (D). 4. **Part (c): Histogram shape analysis.** - The histogram is skewed right (long tail to the right). - It has an outlier (highest bin with very low frequency). - It is asymmetric. - Therefore, answer is all of the above (D). **Final answers:** - (a) B - (b) D - (c) D