Mathematics

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.  A. 0.16 B. 5.3 C. 5.6% D. 16%


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Clayton1

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You can solve this by using the bigger number (3km) as the divisor and the smaller number (2.84km) as the dividend ( 2.84km divided by 3km) to find the number 0.946666 and so on. This is the decimal part of how much David had right and we need to find part he had wrong so you subtract this from a whole (1) and get 0.05333334. Now you can convert this to percent by moving the decimal two spot to the right to get 5.333334%, rounding to the nearest tenth gets you 5.3% (Answer B)

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