Answered

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A student was measuring water in a graduated cylinder. The student read the amount of water at 20 ml. The actual amount of water in the graduated cylinder was 17 ml, What is the approximate percent error?

Sagot :

Recall that percent error is given by the formula:

[tex]|\frac{real\text{ value-measured value}}{\text{real value}}|\cdot100[/tex]

Therefore in our case this becomes:

[tex]|\frac{17-20}{17}|\cdot100=|\frac{3}{17}|\cdot100\approx17.65[/tex]

Therefore the rounded (to two decimals) answer is : 17.65 %

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