A statement that 'the ice-cream sales figure will be approximately 20% of the ice-cream volume sold' translates to
S ~ 0.2 V
where S is sales figure and V is volume sold. So, what does your fit look like? If it were
\(\displaystyle \overset{\wedge}{S}\, =\, 0.21\, V +0.01\)
and V was always large compared to .01/.21, say greater than 1, then yes S ~ 0.2 V would be a good estimate. If it were a different function which might appear not to indicate that at all, then depending on just what is wanted you should
(1) compute some kind of confidence interval for your fit if what you want is an estimation for present data
or
(2) compute some kind of prediction interval estimate if what you want is an estimation for future data. Note that this will probably involve computing some sort of 'goodness of fit' values such as an R2, Std. Est., confidence interval, etc. You will also need to obtain/assume a model for the 'errors' associated with the data. That is the actual data would be something like
yi = a + b xi + Ni
where the Ni are from some probability density function, i.e. possibly a normal distribution with mean \(\displaystyle \mu\) and standard deviation \(\displaystyle \sigma\) which you would need to know or estimate.
