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3 years ago

8

David measured the distance from his house to a soccer field as 3 km. The actual distance from his house to the soccer field is

2.84 km.
Solve for the approximate percent error in David’s measurement. Round to the nearest tenth of a percent, if necessary. A. 0.16% B. 5.3% C. 5.6% D. 16%

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

The percentage error is given by:
(error/actual value) x 100
= [(3-2.84)/2.84] x 100
= 5.6%
The answer is C

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