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^{2}. Why are the last two digits of this number in parenthesis? Is this the same thing as saying (0.510 998 9461 ± 0.000 000 0031) MeV/c

^{2}.

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- Thread starter Mason Smith
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- #1

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- #2

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± 0.000 000 0031

No, it does not mean that. It means that it is approximately equal to 0.510 998 946131 but the last two decimal places are uncertain.

- #3

Lord Jestocost

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If, for example,

- #4

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OK if what @Lord Jestocost said is true then I was wrong.

- #5

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This helped a lot. Thank you for the reference.Use of concise notation(from https://physics.nist.gov/cgi-bin/cuu/Info/Constants/definitions.html)

If, for example,y= 1 234.567 89 U andu(y) = 0.000 11 U, where U is the unit ofy, thenY= (1 234.567 89 ± 0.000 11) U. A more concise form of this expression, and one that is in common use, isY= 1 234.567 89(11) U, where it understood that the number in parentheses is the numerical value of the standard uncertainty referred to the corresponding last digits of the quoted result.

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