deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 0562e4b..80126c5 100644  --- a/untitled.tex  +++ b/untitled.tex 
...
\textbf{Sample Data and Expected Results}  Based on data from Reserve Bank Coinage Specifications[1], the densities for 10 cent, 20 cent, 50 cent, 1 dollar and 2 dollar coins should come out as 6328, 6901, 6113, 7027, 6715 kg/m3 respectively. So values should roughly fall within 6000-7000 kg/m3 (Note: I did measurements with the coins myself on Thursday. Diameters and masses are pretty consistent between testing one coin and five coins, but thickness isn’t - it’s very dependant on exactly where you test it. Students will probably need to measure the thickness of the coin at the *rim* to get a density in the range above. I took thicknesses from the center of each coin, and the calculated densities were closer to 7500kg/m3 - Alex)\\  \begin{table}   \begin{tabular}{ c c c c c c }  & 10c & 20c & 50c & $1 & $2 \\   Thickness($\pm0.01$mm) & 1.37 & 1.48 & 1.39 & 2.59 & 2.37 \\   Diameter($\pm0.01$mm) & 20.48 & 21.70 & 24.76 & 22.94 & 26.55$\pm$0.05 \\   Mass($\pm0.1$g) & 3.4 & 4.1 & 5.1 & 8.0 & 9.8 \\   \end{tabular}   \caption{Single coin measurements}   \end{table}  \phantom{1}\\  For the last question of part 1, the key thing for students to realise is that if 10 coins were used then each relative uncertainty would decrease by a factor of 10, which would ultimately propagate to 10 times the precision. Writing out the whole derivation would probably take too long, so I wouldn’t be picky about that.\\  \phantom{1}\\