Mistake in calculating the last time of battery (Ampere hour problem)
You made the common mistake we all make now and again and inverted the ratio.

It should be
$$\frac{20\space\text{h}\times\frac{1}{20}\space\text{A}}{0.088\space\text{A}}\approx11.36\space\text{h}\tag1$$
Or you could simply have divided \$1\text{Ah }/\text{ } 0.088\text{ A} = 11.36\text{ h}\$
(Note the top part of the corrected equation just undoes what you did to calculate the 20hrs in the first place.)
Whether you actually get those numbers is however dependent on a number of other factors that are way beyond the simple equations and out of scope for an answer here. Unfortunately those &quot;other factors&quot; are rather less well defined mathematically and include things that have more to do with chemistry, and rates of exothermal reactions, than electronics. Suffice it to say, unless you are using extremely low currents or extremely large currents, the formula above is close enough.
Variance battery to battery can be large though. It really depends on the quality of the manufacturer and the history of what happened to the thing before you received it. If it's from a fully automated ISO company and fresh from the factory and not made on a Friday afternoon, the Ah number is probably within about 10% correct. If it's some outfit in Lower Bungholia that job lots manufacture out to local &quot;house-wives&quot; and ships them around in donkey carts... the numbers mean nothing.