 To use the weighted arithmetic mean, the densities would have to be weighted by volume.Applying dimensional analysis to the problem while labelling the mass units by element and making sure that only like element-masses cancel, makes this clear.If one connects two electrical resistors in parallel, one having resistance "x" (e.g., 60 Ω) and one having resistance "y" (e.g., 40 Ω), then the effect is the same as if one had used two resistors with the same resistance, both equal to the harmonic mean of "x" and "y" (48 Ω)