# Tag Info

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for simplicity of notation say $P = \frac{X - N}{X + N}$ given $\delta X$ is the uncertainty in X and $\delta N$ is the uncertainty in N then $\delta (X - N)$ = $\delta (X + N) = \sqrt {\delta ^2X + \delta ^2N}$ and therefore: $\delta P = P \sqrt{(\frac{\sqrt {\delta ^2X + \delta ^2N}}{X - N})^2 + (\frac{\sqrt {\delta ^2X + \delta ^2N}}{X + N})^2}$ ...

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What happens at higher temperatures? Is there an upper asymptote? There appears to be an inflection point around temp=53. This is visible on the log-linear plot. I would try fitting a logistic function http://en.wikipedia.org/wiki/Generalised_logistic_curve as the relationship looks more sigmoidal than exponential (assuming the data is accurate).

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I think you have two distinct thermally activated processes going on here. If I do a log-linear plot of your data I get: It looks to me as if below 45°C the points lie on a straight line and above 45°C the points lie on a steeper straight line. If I do a linear regression of the points below 45°C and above 45°C I get the fits: Below 45°C:  S = 0.771 ...

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$k$ 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 ...

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Benford's law is pretty cool. It states that, for many sets of data, a leading digit of n has a probability of $Pr(n) = log_{10}(1+1(n))$ Plugging in our n values we find that we can expect low values of n to have a higher probability of being our leading digit. The most (initially) boggling thing is that our $Pr(1) = .301$ stays independent of units. If ...

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