a = amount invested at 7%
b = amount invested at 9%
we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that
since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.
![\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20a%20%2B%20b%20%3D%2036000%5C%5C%5C%5C%200.07a%2B0.09b%3D2920%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%201st%20equation%7D%7D%7Ba%20%2B%20b%20%3D%2036000%5Cimplies%20%5Cunderline%7Bb%20%3D%2036000-a%7D%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B0.07a~~%20%2B%20~~0.09%28%5Cunderline%7B36000-a%7D%29~~%20%3D%20~~2920%7D%20%5C%5C%5C%5C%5C%5C%200.07a%2B3240-0.09a%3D2920%5Cimplies%203240-0.02a%3D2920%5Cimplies%20-0.02a%3D-320%20%5C%5C%5C%5C%5C%5C%20a%3D%5Ccfrac%7B-320%7D%7B-0.02%7D%5Cimplies%20%5Cboxed%7Ba%3D16000%7D~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B36000~~%20-%20~~16000%7D%7B20000%3Db%7D%7D)
I think -8 -64 but srry if wrong
Answer:
at first we put the numbers in order from least to greatest
2 , 6 , 6 , 7 , 8 , 9
1st quartile = 6
median = (6+7)/2 = 6.5
3rd quartile = 8
Since we know the x and y, you would divide 480 by 15 and 32 miles per hour. Then you would multiply 7.2 by 32 and get 230.4. Your trying to find with wind and against wind so just times that by two and get 460.8. I think
