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Old Nov 5, 2012, 10:45 PM   #26
AustinIllini
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Quote:
Originally Posted by Signal-11 View Post
The three elipses following the three nines is another way of stating an infinitely repeating series of 9s.

As dukebound stated, 54.999... rounds to 60.

Just a bit of math trolling.
got it. totally blew past the ellipses.
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Old Nov 5, 2012, 10:46 PM   #27
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Originally Posted by dukebound85 View Post
Granted usually more precision is needed than 1 sig fig in practice
More or less what I'm getting at.
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Old Nov 6, 2012, 09:17 AM   #28
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58.629 rounded to one significant figure is 6 * 10^1. If you write 60 you have two significant figures.

(Trust me, I'm an engineer )
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Old Nov 6, 2012, 09:19 AM   #29
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Quote:
Originally Posted by lewis82 View Post
58.629 rounded to one significant figure is 6 * 10^1. If you write 60 you have two significant figures.

(Trust me, I'm an engineer )
No you don't. You have one and one indeterminate.
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Old Nov 6, 2012, 10:26 AM   #30
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Quote:
Originally Posted by lewis82 View Post
58.629 rounded to one significant figure is 6 * 10^1. If you write 60 you have two significant figures.

(Trust me, I'm an engineer )
I'm also an engineer.

60 is one significant figure.

60. is two significant figures.

One's as good as the other, both in theory and in practice, as far as I'm concerned.
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Old Nov 6, 2012, 11:36 AM   #31
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60 ... or 6 x 10^1
Easier to convert 58.629 to power of 10, 5.8629*10^1 , and take 1 sig fig from there.
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Old Nov 6, 2012, 06:06 PM   #32
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Originally Posted by flopticalcube View Post
No you don't. You have one and one indeterminate.
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Originally Posted by Tomorrow View Post
I'm also an engineer.

60 is one significant figure.

60. is two significant figures.

One's as good as the other, both in theory and in practice, as far as I'm concerned.
Writing the 0 signifies that it is exactly 60, that is, that the unit value is known and is zero. On the other hand, 6 * 10^1 means that the value is in the order of magnitude of 60, and that the unit value is unknown.

Subtle difference but when one needs to avoid any ambiguity, this is the way to do it. I do agree that, most of the time, writing 60 is perfectly acceptable.
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Old Nov 6, 2012, 07:40 PM   #33
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Quote:
Originally Posted by lewis82 View Post
Writing the 0 signifies that it is exactly 60, that is, that the unit value is known and is zero. On the other hand, 6 * 10^1 means that the value is in the order of magnitude of 60, and that the unit value is unknown.

Subtle difference but when one needs to avoid any ambiguity, this is the way to do it. I do agree that, most of the time, writing 60 is perfectly acceptable.
Incorrect. The best way to accurately represent 60 is 6.0 X 10^1. Trailing zeroes are indeterminate.

http://www.ruf.rice.edu/~kekule/Sign...gureRules1.pdf

Quote:
Zero Type #3: trailing zeros in a whole number.
200 is considered to have only ONE significant figure while 25,000 has two.
This is based on the way each number is written. When whole number are written as above, the
zeros, BY DEFINITION, did not require a measurement decision, thus they are not significant.
However, it is entirely possible that 200 really does have two or three significant figures. If it
does, it will be written in a different manner than 200.

Typically, scientific notation is used for this purpose.
http://en.wikipedia.org/wiki/Significant_figures

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The significance of trailing zeros in a number not containing a decimal point can be ambiguous. For example, it may not always be clear if a number like 1300 is precise to the nearest unit (and just happens coincidentally to be an exact multiple of a hundred) or if it is only shown to the nearest hundred due to rounding or uncertainty. Various conventions exist to address this issue:
A bar may be placed over the last significant figure; any trailing zeros following this are insignificant. For example, 1300 has three significant figures (and hence indicates that the number is precise to the nearest ten).
The last significant figure of a number may be underlined; for example, "2000" has two significant figures.
A decimal point may be placed after the number; for example "100." indicates specifically that three significant figures are meant.[2]
In the combination of a number and a unit of measurement the ambiguity can be avoided by choosing a suitable unit prefix. For example, the number of significant figures in a mass specified as 1300 g is ambiguous, while in a mass of 13 h*g or 1.3 kg it is not.
However, these conventions are not universally used, and it is often necessary to determine from context whether such trailing zeros are intended to be significant. If all else fails, the level of rounding can be specified explicitly. The abbreviation s.f. is sometimes used, for example "20 000 to 2 s.f." or "20 000 (2 sf)". Alternatively, the uncertainty can be stated separately and explicitly, as in 20 000 1%, so that significant-figures rules do not apply.
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Old Nov 8, 2012, 06:42 AM   #34
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We have a bunch of engineers and no one can agree how to round a number, imagine that.
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Old Nov 8, 2012, 11:01 AM   #35
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I agree with those that say 4.99999.... rounds to 10.

I also agree that 58.2349234 rounds to 60, correct to 1 significant figure. (I don't know why you'd ever need only 1 significant figure...but..yeah).

65 rounds to 60, programming wise. 55 rounds to 60, programming wise. (Even numbers round down, odd rounds up?)
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Old Nov 8, 2012, 11:23 AM   #36
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Can someone explain a practical use of significant figures? I kind of get the fact that they are used to remove slight inaccurate measurements, but not quite sure to what degree they are used.

For instance, if I measure the leg of a table to 42.25" on my tape measure, I'm not going to round it to 40 to cut the other legs. How do you decide how many sig figs to use?
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Old Nov 8, 2012, 12:51 PM   #37
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Originally Posted by samiwas View Post
Can someone explain a practical use of significant figures? I kind of get the fact that they are used to remove slight inaccurate measurements, but not quite sure to what degree they are used.

For instance, if I measure the leg of a table to 42.25" on my tape measure, I'm not going to round it to 40 to cut the other legs. How do you decide how many sig figs to use?
In my weights and measurements lab, it was explained to us that it has to do with the least counts of your measuring apparatus.

For example, if your scale has a least count of 0.01 gram, then the amount of measuring error is +/- 0.01 gram. Multiplied out with other measurements you may have taken, the larger the possible error in any of them (based on least counts), the larger the possible error in your final answer. So measuring mass to within 0.01 gram, length to within 1 mm, and time to 0.1 second is all well and good, but if you introduce some measurement with an apparatus with a large least count - say, you measure a force to within +/- 100 N - then the accuracy of your other measures is relatively meaningless by comparison, and you use whichever measurement has the fewest number of significant figures.
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Old Nov 8, 2012, 12:58 PM   #38
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Quote:
Originally Posted by samiwas View Post
Can someone explain a practical use of significant figures? I kind of get the fact that they are used to remove slight inaccurate measurements, but not quite sure to what degree they are used.

For instance, if I measure the leg of a table to 42.25" on my tape measure, I'm not going to round it to 40 to cut the other legs. How do you decide how many sig figs to use?
Significant figures is required when you are dealing with instruments that have a certain precision.

For instance, if you are involving measurements that are only significant to 4 decimal points, such as 42.25" then any manipulation of that number in calculations should never exceed the significant digits that were initially measured with as they are meaningless

Much akin to how (3.5*2.3)/1.8 =4.5 and NOT the 4.47222 the calculator will show if your level of precision is only to two significant figures, as it is as the inputs are only precise to 2 sig figs

However (3.50000*2.30000)/1.80000 = 4.47222 since the original numbers are designated to be precise out to 6 sig figs

and (3.5*2.30000)/1.80000 = 4.5 as your lowest value is only known to 2 significant digits so the resultant calculation can only be known to 2 significant digits

Last edited by dukebound85; Nov 8, 2012 at 01:08 PM.
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