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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. What is the approximate percent error in David’s measurement? Round to the nearest tenth of a percent, if necessary. I think the answer is A A. 0.16% B. 5.3% C. 5.6% D. 16%
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((3-2.84)/3)*100=?
=5.333333
change / actual as a percentage
triciaal is telling you that you should divide the by the *true distance* 2.84 not by the measurement 3
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