# Precision Question

I am measuring the position of photogates in a free fall lab experiment.

The question is:
What is the error in s in terms of reading error? Now add in the error in the metal scale, a precision of one part in 4000, in quadrature to give an expression for the total error in s.

My reading error is +/- 0.02cm.
I'm confused about 'adding in the error in the metal scale, 1 part in 4000'.
Does this mean 1/4000 = 0.00025?

If so then would my total error would be $$\sqrt{0.00025^2 + 0.02^2}$$ =0.02 ... I don't think this makes much sense

Any help would be greatly appreciated.
Thanks!