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November 28, 2014

November 28, 2014

Posted by **des. student!** on Thursday, September 13, 2007 at 8:26pm.

(6.17 x 10-1 + 4.9 x 10-2) x (3.95 x 10-2 + 6.454 x 10-3)

i know how to solve for the answer, i just dont know the correct number of sig figs! please help!!!!

- math? -
**Count Iblis**, Thursday, September 13, 2007 at 10:23pm4.9*10^(-2) is known to two significant figures, so the answer should be given in to significant figures.

Note that this is only a rule of thumb. In general you need to work as follows. You have some function like:

f(x1,x2,x3,x4)

E.g., in this case

f(x1,x2,x3,x4) = (x1 + x2)x(x3 + x4)

and x1, x2, x3 and x4 are known to finite precision. If x1 = 6.17 x 10(-1) then that essentially means that x1 could be anything in the interval

6.165x 10(-1) to 6.175x 10(-1)

You can thus say that:

x1 = 6.17 x 10(-1) ± 0.005 x 10^(-1)

0.005 x 10^(-1) is thus the error in x1, let's call this sigma1. Similarly you can find sigma2 ...sigma4, the errors in the other variables.

To compute the error in f, you proceed as follows. You just change the value for each variable, one by one, by its error and look at by how much the value of the function changes:

sigmaf1 = f(x1+sigma1,x2,x3,x4) -

f(x1,x2,x3,x4)

sigmaf2 = f(x1,x2+sigma2,x3,x4) -

f(x1,x2,x3,x4)

etc.

The error in f is then given by:

sigmaf = sqrt[sigmaf1^2 + sigmaf2^2 + sigmaf3^2 + sigmaf4^2]

From the value of sigmaf you can then see how many significant digits of f can be given.

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