Reading this document, when measuring a thing we have relative error = absolute error value of thing measured .Let's say we have a measurement of the height of an object as 45.2m with error of ± 0.05 m.At that point our relative error would be 0.05 m 45.2 m .However, if a person who measures heights always rounds his measurement to three significant digits would it change the relative error?I am not sure if this is correct, but in other words would the answer 45.2m+0.05m=45.25m rounded to 45.3m as the approximate value. After that would th e absolute error be 0.1m, so that the relative error would be 0.1 45.2 ?

Question

Reading this document, when measuring a thing we have  relative error  =      absolute error    value of thing measured    .Let&#039;s say we have a measurement of the height of an object as 45.2m with error of   ±  0.05  m.At that point our relative error would be             0.05      m              45.2      m      .However, if a person who measures heights always rounds his measurement to three significant digits would it change the relative error?I am not sure if this is correct, but in other words would the answer 45.2m+0.05m=45.25m rounded to 45.3m as the approximate value. After that would th e absolute error be 0.1m, so that the relative error would be       0.1    45.2  ?

Donavan Mack · Accepted Answer

Three significant digit cancellation wont have an effect in this question and the answer will be same. The answer proposed in the question is the right one as 0.05/45.2 relative error.

Reading this document, when measuring a thing we have <mtext>relative error</mtext> =

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