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forsale [732]
3 years ago
12

He measured the pool’s dimensions to be 15 feet long by 35 feet wide. The actual dimensions of the pool are 14.5 feet by 34.75 f

eet. Find Jack’s percent error to the nearest percent.
Mathematics
1 answer:
mote1985 [20]3 years ago
3 0

Answer:

Percent error = -4.02 (Approx.)

Step-by-step explanation:

Given:

Estimated pool dimension = 15 ft long and 35 ft wide

Actual pool dimension = 14.5 ft long and 34.75 ft wide

Find:

Percent error

Computation:

Area of Estimated pool = 15 x 35

Area of Estimated pool = 525 ft²

Area of Actual pool = 14.5 x 34.75

Area of Actual pool = 508.875 ft²

Percent error = [(Actual - Estimated) / Estimated]100

Percent error = [(508.875 - 525) / 525]100

Percent error = [(-21.125) / 525]100

Percent error = -4.02 (Approx.)

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