1 cup = 8 fl oz.
So then we can set up a ratio of
![\sf \frac{fluid ~ounces}{cups}](https://tex.z-dn.net/?f=%5Csf%20%5Cfrac%7Bfluid%20~ounces%7D%7Bcups%7D%20)
So we get:
![\sf\frac{8~ fl. oz.}{1 ~cup} = \frac{22 ~fl. oz.}{x~ cups}](https://tex.z-dn.net/?f=%20%5Csf%5Cfrac%7B8~%20fl.%20oz.%7D%7B1%20~cup%7D%20%3D%20%20%5Cfrac%7B22%20~fl.%20oz.%7D%7Bx~%20cups%7D%20)
We cross multiply and we try to solve for x, so:
![\sf 8~fl.~oz. \times x ~cups = 1~cup \times 22~fl.~oz.](https://tex.z-dn.net/?f=%5Csf%208~fl.~oz.%20%5Ctimes%20x%20~cups%20%3D%201~cup%20%5Ctimes%2022~fl.~oz.)
![\sf x~cups = \frac{22~fl.oz.}{8~fl.oz.}](https://tex.z-dn.net/?f=%5Csf%20x~cups%20%3D%20%20%5Cfrac%7B22~fl.oz.%7D%7B8~fl.oz.%7D%20)
And the fl. oz. cancel out, and we are left with:
<u>So 22 fl. oz. equals to 2 3/4 cups.</u>
We have q
consecutive integers in a set and q is odd.
If 120 is the middle number, then
there will be (q-1)/2 numbers
both before
and after 120.
Therefore, the largest number will
be (q-1)/2+120.
Answer is: -17+6+(-3)=-14