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A student measured the string as 2.43 m long. The teacher said it was actually 2.12 m long. What was the student's percent error?
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20 votes
A student measured the string as 2.43 m long. The teacher said it was actually 2.12 m long.

What was the student's percent error?

User John Cast
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1 Answer

10 votes
10 votes
Percent error = Actual observed value - expected value / expected value.
= 2.43-2.12/2.12 * 100 = 14.6226% error
User ChrisThompson
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