Recent content by Dennisl

Oh I didn't realize those were d's. so it is \sqrt{\left(\frac{1}{2\sqrt{5}}\times.02\right)^{2}} I get 2.2360±0.0044 as well. I see how it works. Thank you for your help!

so for my example point 5.00 +- .02 then y=\sqrt{\left(\frac{\sqrt{5}}{5}\times\frac{.02}{5.00}\right)^{2} = 2.2361 \pm .0018 (ignoring sig figs)?

Homework Statement I'm doing a lab and I'm having difficulty calculating error for a linearized set of data. I need to find the error for a set of data that has been curved to a square root. for example my x= 5.00 +- .02 becomes sqrt(5) but what is the error? Homework Equations The...