Applying Sig Figs to Complex Calculations

[firstwave]

I understand that if your doing multiplication, round to the least amount of sig figs. Addition: Round to the least amount of decimal places.

What about a question with both multiplication, division and addition, subtraction?

Such as

(5.2 x 10^4 + 4.2 x 10^-2)/ 3.6 x 10^2

How would I apply sig figs on that?

Thx in advance

 

[Sirus]

\frac{52000+0.042}{360}=\frac{52000.042}{360}\approx 144.4445611\approx 144

That is how I would do it. 3 significant digits in the 360, so the same for the answer, and don't round anything until the end.

 

[endeavour]

\frac{52000+0.042}{360}=\frac{52000.042}{360}\approx 144.4445611\approx 144

That is how I would do it. 3 significant digits in the 360, so the same for the answer, and don't round anything until the end.

I think it's about context too. If the question was about a length measurement taken (just say) and the value was 360mm, then shouldn't that should be taken to 2 sig.fig (unless they actually say it was measured to 3 sigfig)?

 

[Sirus]

I'm assuming the instrument used had millimeter markings. Good point, though.

 

[firstwave]

thx guys :D

I think 360 is 2 sig figs

360.0 = 4 sig figs

 

[Doc Al]

I think 360 is 2 sig figs

When written as in your first post (3.6 x 10^2) it has 2 sig figs for sure; 3 sig figs would be written as 3.60 x 10^2

 

[Sirus]

Oh boy, big mistake on my part. Corrected:

\frac{52000+0.042}{360}=\frac{52000.042}{360}\approx 144.4445611\approx 1.4\times{10^{2}}

Firstwave, do the calculation, then look over to see the least number of significant digits in the original data. Sorry for misleading you earlier; my mistake.

 

[firstwave]

ok thanks all

 

[Diane_]

Oh boy, big mistake on my part. Corrected:

\frac{52000+0.042}{360}=\frac{52000.042}{360}\approx 144.4445611\approx 1.4\times{10^{2}}

Firstwave, do the calculation, then look over to see the least number of significant digits in the original data. Sorry for misleading you earlier; my mistake.

Sirius' solution is correct, but the pedantic side of me insists I point out that the numerator of that expression has only two significant digits, despite being written as though it had eight. Since you don't really know what the digits are in the "52000" after the 2, adding the .042 makes no significant difference in the number.