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How many significant figures are there in $5~\mathrm{ft}\ 10~\mathrm{in}$. I know that $5~\mathrm{ft}\ 10~\mathrm{in}= 5.8\bar{3}~\mathrm{ft}=70~\mathrm{in}$, so I am not sure exactly how much significant digits are in this number.

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Significant figures is a rule of thumb, to avoid giving overly precise results when the input doesn't warrant it.

In actual scientific calculations - where keeping track of the precision matters - significant figures aren't used. Instead, explicit error ranges (confidence intervals) are used. Notation of these varies, but often takes the form of a plus-minus format (14.5±0.3 m). The concept of significant figures uses implicit error ranges, conventionally set to one half of the last significant digit.

When error ranges are explicitly given, the error of a calculation result is found with formal propagation of uncertainty, rather than the rule-of-thumb significant figure rules. Sometimes this is based on a model of the distribution of error around the central value (for example, assuming that the error is normally distributed around the central value with a given standard deviation), but most commonly the error is taken as the minimum and maximum of an acceptable range of values around the main measurement. Under this scheme, the propagation of uncertainty often takes the form of the "crank three times" method, whereby you run the calculation once for the central value, once for the minimum value, and once for the maximum value.

So if you were doing a calculation like $\pu{14.5\pm0.3 m} - \pu{1.5\pm0.2 m}$, you would calculate once for the central value ($\pu{14.5 m - 1.5 m = 13 m}$), once to give you the smallest value you could get with the range ($\pu{14.2 m - 1.7 m = 12.5 m}$), and once to give the largest value your could get with the range ($\pu{14.8 m - 1.3 m = 14.5 m}$). You'd then convert this into a range, normally rounding both values such that the error value only has 1-2 significant figures: $\pu{13\pm0.5 m}$

It turns out that the rules for significant figure calculations are just a quick-and-dirty rule of thumb of doing more-or-less the same thing. I'd encourage you to try it out with a number of examples. If you assume that the error in each quantity is one half of the last significant digit, and you do the crank three times error propagation, you should come up with a result that has an error that is about one half the last significant digit of the answer by the significant figures approach - give or take a digit.

So really what you want to do is to convert the measurement of $\pu{5 ft 10 in}$. into one which has an explicit error associated with it. For this measurement, I'm guessing you'd have an error of $\pu{\pm0.5 in}$. - I say this because rounding to the nearest 10 inches seems rather silly to me. (Feel free to work the calculations below with different assumption, say with ±5 in., if you disagree.) So you have a measurement of $\pu{5 ft 10 in. \pm 0.5 in}$ in other words $\pu{5 ft} (\text{exact}) + 10\pm\pu{0.5 in}. = \pu{5 ft} (\text{exact}) \times \pu{12 in./ft }(\text{exact}) + \pu{10\pm0.5 in.} = \pu{70\pm0.5 in.}$ Note that is the same as $70$. in with two significant digits.

Since we have explicit error bars, we can use the crank three times method to convert it to feet: $70~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.8\bar{3}~\mathrm{ft}$ and $69.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.791\bar{6}~\mathrm{ft}$ and $70.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.875~\mathrm{ft}$. This is $5.8\bar{3}±0.041\bar{6}~\mathrm{ft}$. Convention holds that we round the error to one or two significant figures, and also round the main value to the same decimal place: $5.83±0.04~\mathrm{ft}$. -- This is basically what you'd have from $\pu{5.8 ft}$ with two significant digits.

The upshot is that you'd never use $\pu{5 ft 10 in.}$ in a calculation. You'll want to convert it to a single consistent unit first. When you do that, keep in mind the actual error associated with the measurement, and use that error as a gauge at how precise the converted quantity is.

(Note that if this is a question on a test or on a homework, ask your teacher how they think you should get the "correct" answer. When dealing with rule-of-thumb things like significant figures, intelligent people can vary in their interpretations. This means an answer valid in one context can be marked as wrong on the answer key because it doesn't match the formalism the teacher is using.)

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  1. All non-zero digits are significant(62 has 2 sig fig)
  2. In a number without a decimal point, only zeros between non-zero digits are significant(206's 0 is significant).
  3. In a number with a decimal point, all zeros to the right of the first non-zero digits are significant(eg. 15.30 the zero is significant but 015.3 the zero is not significant. 15.30 has 4 sig fig but 015.3 only has 3).
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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ – Martin - マーチン Aug 12 '15 at 8:37
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    $\begingroup$ This is true but doesn't address the OP's problem which is specifically about significant figures in non-SI unit systems. $\endgroup$ – bon Aug 12 '15 at 10:16
  • $\begingroup$ Express the specified value in inches, as you have done (since that is a more precise measure of length than feet) and then count the significant figures. $\endgroup$ – iad22agp Aug 12 '15 at 12:26

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