SOLUTION: a bricklayer measured the length of his wall to be 4.10m, if the actual length is 4.25m, what is his percentage error

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Question 982906: a bricklayer measured the length of his wall to be 4.10m, if the actual length is 4.25m, what is his percentage error
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Error=((Actual-Measured)/Actual)*100
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E=%28%284.25-4.1%29%2F4.25%29%2A100=3.5
3.5%