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A student measured a strip of aluminum as 1.25m long. The teacher said it was actually 2.12m long. What was the student's percent error?

Respuesta :

Answer:

41.03 %

Explanation:

Given that,

The actual length of a strip of aluminum is 2.12 m

Observed length of a strip of aluminum is 1.25 m

We need to find the student's percent error. The percentage error on any value is given by :

[tex]\%=\dfrac{\text{observed-actual}}{\text{actual}}\times 100[/tex]

Putting all values, we get :

[tex]\%=\dfrac{2.12-1.25}{2.12}\times 100\\\\\%=41.03\ \%[/tex]

So, the percent error of the student is 41.03 %.