1. **State the problem:** Coach Leo scheduled \(\frac{2}{5}\) hours for dribbling drills and \(\frac{1}{4}\) hour for team scrimmage. The gym is available for 1 hour. We need to determine if the total scheduled time fits within 1 hour. 2. **Add the scheduled times:** Calculate the total time as the sum of the two parts: $$\frac{2}{5} + \frac{1}{4}$$ 3. **Find a common denominator:** The denominators are 5 and 4. The least common denominator is 20. Convert each fraction: $$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$$ $$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$$ 4. **Add the converted fractions:** $$\frac{8}{20} + \frac{5}{20} = \frac{13}{20}$$ 5. **Compare total time to gym availability:** \(\frac{13}{20} = 0.65\) hours, which is less than 1 hour. 6. **Conclusion:** The team will finish on time because \(0.65 < 1\) hour. Coach Leo does not need to adjust the schedule.