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 need to add up the lengths of all the jumps. Given: - Jump A: \(12 \frac{1}{4}\) feet - Jump B: \(16 \frac{1}{2}\) feet - Jump C: \(17 \frac{5}{8}\) feet - Jump D: \(18 \frac{1}{4}\) feet Let's calculate the total distance by adding up the lengths of all the jumps. Calculate the value by following steps: - step0: Calculate: \(12\frac{1}{4}+16\frac{1}{2}+17\frac{5}{8}+18\frac{1}{4}\) - step1: Convert the expressions: \(\frac{49}{4}+\frac{33}{2}+\frac{141}{8}+\frac{73}{4}\) - step2: Reduce fractions to a common denominator: \(\frac{49\times 2}{4\times 2}+\frac{33\times 2\times 2}{2\times 2\times 2}+\frac{141}{8}+\frac{73\times 2}{4\times 2}\) - step3: Multiply the numbers: \(\frac{49\times 2}{8}+\frac{33\times 2\times 2}{2\times 2\times 2}+\frac{141}{8}+\frac{73\times 2}{4\times 2}\) - step4: Multiply the terms: \(\frac{49\times 2}{8}+\frac{33\times 2\times 2}{8}+\frac{141}{8}+\frac{73\times 2}{4\times 2}\) - step5: Multiply the numbers: \(\frac{49\times 2}{8}+\frac{33\times 2\times 2}{8}+\frac{141}{8}+\frac{73\times 2}{8}\) - step6: Transform the expression: \(\frac{49\times 2+33\times 2\times 2+141+73\times 2}{8}\) - step7: Multiply the numbers: \(\frac{98+33\times 2\times 2+141+73\times 2}{8}\) - step8: Multiply the terms: \(\frac{98+132+141+73\times 2}{8}\) - step9: Multiply the numbers: \(\frac{98+132+141+146}{8}\) - step10: Add the numbers: \(\frac{517}{8}\) The total distance of the jumps is \(64\frac{5}{8}\) feet or 64.625 feet.