Significant figures

  • #1
LCSphysicist
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Homework Statement:
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Relevant Equations:
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I am having a little trouble to write in the right way a number:

I did a calc in which the uncertainty was 150, while the value with physics meaning was about to 1427.

I am trying to figure out what is the right way to write it.
Now, since the uncertainty can not have more than two significant figure, my guess would be:

$$1.4*10^3 \pm 1.5*10^2$$

My doubt is about the first number, what are the rules about it? The same number of significant figures that uncertainty has? But normally if the uncertainty is, let's say, "0.003" and the value measured "154.3464". I would write $$154.346 \pm 0.003$$ and i am almost sure this is right, but it does not have the same quantity of significant figures. I am confused
 

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  • #2
haruspex
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The last digits of the value and the uncertainty should align, so maybe

$$1.43*10^3 \pm 1.5*10^2$$
 
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  • #3
Steve4Physics
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Homework Statement:: .
Relevant Equations:: .

I am having a little trouble to write in the right way a number:

I did a calc in which the uncertainty was 150, while the value with physics meaning was about to 1427.

I am trying to figure out what is the right way to write it.
Now, since the uncertainty can not have more than two significant figure, my guess would be:

$$1.4*10^3 \pm 1.5*10^2$$

My doubt is about the first number, what are the rules about it? The same number of significant figures that uncertainty has? But normally if the uncertainty is, let's say, "0.003" and the value measured "154.3464". I would write $$154.346 \pm 0.003$$ and i am almost sure this is right, but it does not have the same quantity of significant figures. I am confused
Sorry, answer is a bit long.

In standard form, principal value is:
1427 = 1.427*10³.

Uncertainties are generally rounded to one, or sometimes two, significant figures. It depends on your local practice. Here I’d use two. Matching the principal value's power of 10, the uncertainty (150) is expressed as:
0.15*10³.

The initial combined result is (1.427±0.15)x10³. Now round the precision (not the signficant figures) of the principal value to match the precision of the uncertainty. This gives the final answer of:
(1.43±0.15)x10³.

You also could write this as 1430±150.

Of course you can do all that in one step.
___________

If you are required to round uncertainty to one significant figure, the answer would be;
(1.4±0.2)x10³

This could also be written as 1400±200.
___________

If the principal value is 154.3464 and the uncertainty is 0.003, then 154.346±0.003 is correct. You could also write it as (1.54346±0.00003)x10² but that’s rather clumsy.

And don’t forget units!
 

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