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A symmetric error in magnitude is not a symmetric error in flux. Put in the biggest value and smallest value of mc and use the two limits as an asymmetric flux error.

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I do it different way, but it winds up with the same answer as Chris. For products or quotients, I add The percent errors in quadrature. That gives $\sigma_v=[(\sigma_x/x)^2+(\sigma_t/t)^2]^{\frac{1}{2}}(x/t)$ in your case.

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Consider some measurments $x$ and $y$. Say there is some derived quantity $z$ which depends on $x$ and $y$ with some function $z=f(x,y)$, the error on $z$ is $\sigma_z$ and is given by $$\sigma_z=\left[ \left(\frac{\partial z}{\partial x} \right)^2 \sigma^2_x + \left(\frac{\partial z}{\partial y} \right)^2 \sigma^2_y\right]^\frac{1}{2}$$ where $\sigma_x$ is ...

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