What is the total distance of the jumps of this length? \( \begin{array}{ll}\text { A } 12 \frac{1}{4} \text { feet } & \text { B } 16 \frac{1}{2} \text { feet } \\ \text { C } 17 \frac{5}{8} \text { feet } & \text { B } 18 \frac{1}{4} \text { feet }\end{array} \)

To find the total distance of the jumps, we first need to convert each mixed number into an improper fraction or a decimal and then sum them up. 1. \( A = 12 \frac{1}{4} = \frac{49}{4} \) 2. \( B = 16 \frac{1}{2} = \frac{33}{2} = \frac{66}{4} \) 3. \( C = 17 \frac{5}{8} = \frac{141}{8} = \frac{282}{16} = \frac{141}{8} \) 4. \( D = 18 \frac{1}{4} = \frac{73}{4} \) Now, we can convert them all to have a common denominator. The least common multiple (LCM) of the denominators 4, 2, 8 is 8: - \( A = 24/8 \) - \( B = 66/4 = 132/8 \) - \( C = 141/8 \) - \( D = 73/4 = 146/8 \) Now, we sum them up: \[ A + B + C + D = \frac{49}{4} + \frac{66}{4} + \frac{141}{8} + \frac{146}{8} \] First combining: \[ \frac{49 + 66 + 141 + 146}{8} = \frac{402}{8} = 50.25 \] So, the total distance of the jumps is \( 50.25 \) feet.