{(4z-7,z-4,z) |z is any real number} z=-5
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The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.
The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.
To solve the problem, we must:
Know how many 30-minutes have passed during the time period.
1 day = 24 hours
1 hour = 60 minutes = 2 × (30 minutes)
1 day = 24 hours × 2 × (30 minutes)
1 day = 48 × (30 minutes)
Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.
Friday = 48 × = 24
Saturday = 48
Sunday = 48
Monday = 48 × = 36
TOTAL = 24 + 48 + 48 + 36 = 156
Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.
Divide the number of total seconds late by the number of 30-minutes passed.
That is, the number of total seconds late= 468 seconds ÷ 156
= 3 seconds
Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.
To learn more about clock problems visit:
brainly.com/question/27122093.
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