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Working with significant digits is very prone to error, because it can be misleading. It is much better to work with explicit errors. So to rewrite your example with explicit errors: $\left(1.0 \pm 0.5\right) \times 10^3 + 1.00\pm0.05$ We now add the errors quadratically (assuming they are uncorrelated): $\sqrt{(0.5\times ... 0$1000$has an ambiguous number of significant digits. It could have 1, it could have 4. I think it is generally assumed that it has four unless otherwise stated. (For example, I've seen people put a line over the last significant digit, like this:$1\overline000$.) This is why scientific notation is useful. If you are saying that$1000$has$1$... 0 1000 has 4 significant digits as mentioned before, stating that the measured value is between 999.5 and 1000.5. 1.0 has 2 significant digits, staying it is measured within an accuracy of 0.05. Adding the numbers gives you a result within an accuracy of 0.5, so noting the result with 1 decimal digit is nonsense. If your measured value of 1000 indeed has only ... 4 The last digits in 1000 are absolutely significant, they state that you have not 1200, not even 1001 but exactly 1000. In scientific notation, you would write this as$1.000 \times 10^3$. Compare this to$1\times10^3$where you have just one significant digit. Update: consider the example from the question$1\times10^3+1.0$. The first term could be anything ... 0$k$is just the mean of$k_1$and$k_2$No, the best value of k is calculated using a weighted mean, weighting by the reciprocals of the squares of the respective individual uncertainty values. An accurate measurement must contribute more to the best value than an inaccurate measurement. I thought that I would need to square-sum the errors ... 2 NIST has its own more sophisticated guidelines for reporting uncertainty of measurments. http://physics.nist.gov/Pubs/guidelines/TN1297/tn1297s.pdf There is nothing wrong with reporting two digits in an uncertainity, and many peer reviewed journal articles do. This is especially true when the first digit of the uncertainity is 1. If you report +/-1, 1 ... 0 When adding and subtracting, you can only go to the lowest number of decimal places. That is to say, we are dealing with precision and not significant figures when adding/subtracting numbers. If you have two measuring devices and one is accurate to 0.1mm and the other to 1mm, then you cannot definitively state the combined measure to 0.1mm, you can only ... 0 First, can it be written as the following? $$N_A=6.022\,141\,29\times10^{23}\pm0.000\,000\,27\times10^{23} {\rm mol}^{-1}$$ Yes. A cleaner way to write it is $$N_A=6.022\,141\,29(27)\times10^{23}{\rm mol}^{-1}$$ where the$(27)$indicated the uncertainty in the last$N$digits (where$N$is the number of digits inside the parenthesis). If I have a ... 0 Brief Overview: The chemist puts a laser on a sample. The atoms in the sample gain energy and for a few nano seconds it shines a light and returns to ground state. People such as agricultural inspectors use this to indentify any chemi 0 Wigner function is not a true probability distribution as it may possess negative values, particularly when the quantum state has no classical analogue. 0 This was a comment but it got too long.. With the level of information you can provide this is not a physics question at all. 10 kWh over 48hr may be very high or very low consumption, unless you have traced the circuits you don't know what that meter supplies. The fact you run out of hot water tells us nothing useful at all about power consumption, it ... 5 The error would be in the order of 10^-14. This is mathematically similar to the sense of errors you have on your hand watch, caused by mechanical inaccuracy - probably in the range of 1 second per week, or 1 second per year if its a Rolex :) One should note however, that such a very small inaccuracy in time measurement in atomic clocks is perhaps less than ... 5 It means that if the clock begins set to the correct time, then after time$t$the clock will be wrong by no more that$(\pm 10^{-14}) t\$. Or as a physicist would be likely put it $$\frac{\delta t}{t} \le 10^{-14} \,.$$ This kind of expression of "fractional errors" is very common in many fields of quantitative science. Now, to be concrete, a year is ...