Significant Figures: How to Properly Round Numbers

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SUMMARY

The discussion centers on the proper application of significant figures in rounding numbers, specifically the example of adding 37.26972, 25.43, 0.837, 101.22, and 3.1, resulting in 167.85672. The correct rounding to one decimal place is 167.9, as the digit being dropped is 5 and the following digit is non-zero, which necessitates rounding up according to established rules. A referenced website confirms this rounding method, clarifying a common misunderstanding regarding significant figures.

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qwsdcnjkio
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I was reading the Intro to Physics PDF file on this forum about significant figures/digits.

Basically the example used is this; add these numbers up and round off to the correct significant figures/digits:

37.26972+25.43+0.837+101.22+3.1=167.85672=167.9(1 decimal place)

I understand why the answer must be given to 1 decimal place, however, according another web site, http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs6.html,
167.85672 should be rounded down to 167.8 (1 d.p.).

Does anyone know who if anyone is correct?
 
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qwsdcnjkio said:
I understand why the answer must be given to 1 decimal place, however, according another web site, http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs6.html,
167.85672 should be rounded down to 167.8 (1 d.p.).

Does anyone know who if anyone is correct?
You must be misreading that link. It clearly says:
(3) If the digit to be dropped is 5, and if any digit following it is not zero, the last remaining digit is increased by one. For example,

12.51 is rounded to 13.​
In the example you gave (167.85672 rounded to 1 decimal), the digit you are dropping is 5 and fits this rule, so the last remaining digit (8) must be rounded up to 9.
 
i see now, don't think i read (3) properly as you said, thanks for lcearing that up :biggrin:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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