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Mathematics 20 Online
OpenStudy (anonymous):

how do I round 55.0 to two significant figures?

OpenStudy (anonymous):

drop the zero.\[5.5*10^1\]

OpenStudy (anonymous):

In general we know that any non-zero number is significant. So in this case you can just round it to "55" without the decimal. What eseidl said also is true.

OpenStudy (anonymous):

okay thankss!!!

OpenStudy (anonymous):

Do you know what a "significant figure is" though? What makes it "significant"?

OpenStudy (anonymous):

Yes, I know the four rules that one just had me confused because it had a zero at the end

OpenStudy (anonymous):

Not the rules themselves but the point of the rules?

OpenStudy (anonymous):

umm not really..

OpenStudy (anonymous):

Basically it works like this. If a took an ordinary ruler and measured say the thickness of a book which of the following would you believe? 12.1cm or 12.0000389cm? Obviously the first one is reasonable (the ruler and my ability to use it are limited in accuracy, probably to a tenth of a cm or so). The second answer is clearly BS. There is no way I could have measure the book that accurate. So 3 figures are "significant" here (are justifiable)...the second answer would be just 12.0cm. We can't tell 12.0000389 apart from 12.0cm here so those extra values are meaningless.

OpenStudy (anonymous):

Ohhhhh!! Okay I understand!

OpenStudy (anonymous):

so when we add, subtract, multiply, etc measured quantities we can't gain or lose accuracy. We can only keep as much and only as much digits as are actually meaningful (justified). Thats pretty much where the rules come from.

OpenStudy (anonymous):

Cool, you definitely helped a lot! Thanks soo much!

OpenStudy (anonymous):

np :)

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