If student A measures a laptop computer to be 37.2 cm long, and student B measures 36.7 cm, what is the percentage error?

Answer 1

I do not know because I do not know the true value.
However I can give the percentage range from the average of the two.
(37.2+36.7)/2 = 36.95

difference = +/- 0.25

100* .25/36.95 = + or - .68%