92+88+65+79+99=423
423/5=84.6%
5 is from the amount of percentages given
I think the answer is 2 13/20
Or, in decimals: 1.3
41 my friend hope this helps my good friend
![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 729=27^2\\ \qquad (3^3)^2\\ 1000=10^3 \end{cases}\implies 729^{15}+1000\implies ((3^3)^2)^{15}+10^3 \\\\\\ ((3^2)^{15})^3+10^3\implies (3^{30})^3+10^3\implies (3^{30}+10)~~[(3^{30})^2-(3^{30})(10)+10^2] \\\\\\ (3^{30})^3+10^3\implies (3^{30}+10)~~~~[(3^{60})-(3^{30})(10)+10^2]](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20%5C%5C%5C%5C%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20729%3D27%5E2%5C%5C%20%5Cqquad%20%283%5E3%29%5E2%5C%5C%201000%3D10%5E3%20%5Cend%7Bcases%7D%5Cimplies%20729%5E%7B15%7D%2B1000%5Cimplies%20%28%283%5E3%29%5E2%29%5E%7B15%7D%2B10%5E3%20%5C%5C%5C%5C%5C%5C%20%28%283%5E2%29%5E%7B15%7D%29%5E3%2B10%5E3%5Cimplies%20%283%5E%7B30%7D%29%5E3%2B10%5E3%5Cimplies%20%283%5E%7B30%7D%2B10%29~~%5B%283%5E%7B30%7D%29%5E2-%283%5E%7B30%7D%29%2810%29%2B10%5E2%5D%20%5C%5C%5C%5C%5C%5C%20%283%5E%7B30%7D%29%5E3%2B10%5E3%5Cimplies%20%283%5E%7B30%7D%2B10%29~~~~%5B%283%5E%7B60%7D%29-%283%5E%7B30%7D%29%2810%29%2B10%5E2%5D)
now, we could expand them, but there's no need, since it's just factoring.
??? Depends on how many people are there