# How to round in error analysis

Hi, to start with my questions I will show you what I have done so far.

(23.56+/-0.05) km/h x (56.3+/-0.4) h

So I ended up with (1326.428+/-12.234) km
But I know the real answer is (1330+/-10) km
What I dont understand is how I would round to that answer. I do not know what I have to look for to see that I have to round it.

Quantum Defect
Homework Helper
Gold Member
Hi, to start with my questions I will show you what I have done so far.

(23.56+/-0.05) km/h x (56.3+/-0.4) h

So I ended up with (1326.428+/-12.234) km
But I know the real answer is (1330+/-10) km
What I dont understand is how I would round to that answer. I do not know what I have to look for to see that I have to round it.

Basically, all of the extra digits in the error that you have are extraneous. The 0.234 past the decimal is essentially meaningless.

The error will be the square root of the sum of the squares of the individual relative errors.

speed: 0.2% error; time: 0.7% error ==> resultant error = sqrt (0.002^2 + 0.007^2) = 0.0074 --> 0.74%
distance = 1326.4 km (0.74% of this is the error) -- error = 9.8 km ==> which has been rounded to 10 (one sig fig)
So, the answer is 1330 +/- 10 km (you round the answer to have the same precision as the precision in the error)

haruspex
Homework Helper
Gold Member
The error will be the square root of the sum of the squares of the individual relative errors.