Q. The rodeo stared 7 minutes late. It was scheduled to last 3 hours, but it ran 22 minutes longer. If the rodeo ended at 9:59 p.m, at what time did it begin?

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A. Do I muplite 7 by 3 and than add the answer with 22?

To determine the start time of the rodeo, we need to subtract the total duration of the rodeo (including the delay and the extra 22 minutes) from the end time of 9:59 p.m.

To calculate the duration of the rodeo in minutes, we first find the total duration in minutes:

3 hours = 3 * 60 = 180 minutes

Next, we add the 7-minute delay and the additional 22 minutes:

180 minutes (scheduled duration) + 7 minutes (delay) + 22 minutes (extra time) = 209 minutes

Now, subtract the total duration from the end time:

9:59 p.m - 209 minutes = 7:30 p.m

Therefore, the rodeo started at 7:30 p.m.

To determine the start time of the rodeo, you need to subtract the total duration (which includes the delay and the extra 22 minutes) from the ending time.

First, convert the starting delay from minutes to hours. There are 60 minutes in an hour, so dividing 7 minutes by 60 will give you the delay in hours: 7 minutes ÷ 60 = 0.117 hours (rounded to 3 decimal places).

Next, add this delay to the scheduled 3-hour duration of the rodeo: 3 hours + 0.117 hours = 3.117 hours (rounded to 3 decimal places).

Now, subtract the total duration (3.117 hours) from the ending time (9:59 p.m).

9:59 p.m - 3.117 hours = 6:52 p.m.

Therefore, the rodeo started at 6:52 p.m.

I have trouble with the rodeo "stared."

21 + 22 does not give you a time.

The rodeo started 3 hours and 29 minutes before 9:59

9:59 - 3:29 = ?