How uncertainties affect values If I calculate the equivalent resistance of a circuit, for example:
$$1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 = 1/1472 + 1/3260 + 1/5580 \Rightarrow R_{eq} = 858.22\,\Omega$$
And then calculate its uncertainty:
$$\Delta R_{eq} = |\delta 1/\delta R_1| \times \Delta R_1 + |\delta 1/\delta R_2| \times \Delta R_2 + |\delta 1/\delta R_3| \times \Delta R_3 = 0.001\,\Omega$$
Will the uncertainty affect the value $(858.22)$, given that it only has 2 decimal numbers?
 A: You have a few errors.
As @NowIGetToLearnWhatAHeadIs notes in a comment, the uncertainties in the resistors are almost certainly uncorrelated, so you want to add them in quadrature (NB: pdf). So that would give you $$(\Delta(1/R_{eq}))^2 = (\delta(1/R_1))^2 + (\delta(1/R_2))^2 + (\delta(1/R_3))^2$$
In addition, $\delta(1/R_1) \neq \delta(1)/\delta(R_1) = 0$. The general form for the uncertainty of a power is $\delta(x^n) = |nx^{n-1}|\delta(x)$ (see Taylor, Introduction to Error Analysis for a derivation). You can use that here with $n=1$.
The final error is that you need the uncertainties in your resistors in order to do any of this. There are a few ways to get this. If you have the actual resistors, and you're unable to actually measure the resistance, you can read the tolerance using the standard resistor color chart. If you can measure the resistance directly, you should be able to get a pretty precise measurement with even a low-end multimeter. Otherwise,you'll have to rely on whatever uncertainty you're told to use.
To answer your title question directly, an uncertainty of $\Delta R_{eq} = 0.001 \Omega$ will not noticeably impact a value of $R_{eq} = 858.22 \Omega$. Any measurement you made to measure that $R_{eq}$ would not be affected by an uncertainty that small. However, I strongly suspect (without working things out myself) that your uncertainty is much larger than that. 
