Question

Please help me with a math problem. See the attachments.

Answer 1

Let x=5.234±0.0005 and y=5.123±0.0005. Find the percentage error of the difference a=x-y when relative errors δx=δy=0.0001.
Solution: Maximum difference is \(x_{max}-y_{min}=5.2345-5.1225=0.112\)
Minimum difference is \(x_{min}-y_{max}= 5.2335 -5.1235=0.111\)
uncertainty of the difference is 0.112-0.111= 0.001
so, the difference is (5.234 - 5.123 )\(\pm\)0.001 = 0.111 \(\pm\)0.001
then percentage of error is \(\dfrac{0.001}{0.111}*100= 0.9\) %

Answer 2

That's all I know, hope this help.

Answer 3

Thanks a lot.