so we know that 24% of the students buy their lunch at the cafeteria, and 190 students brownbag.
well, 100% - 24% = 76%, so the remainder of the students, the one that is not part of the 24% is 76%, and we know that's 190 of them.
since 190 is 76%, how much is the 24%?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 190&76\\ x&24 \end{array}\implies \cfrac{190}{x}=\cfrac{76}{24}\implies 4560=76x \\\\\\ \cfrac{4560}{76}=x\implies 60=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{total amount of students}}{190+60\implies 250}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%20190%2676%5C%5C%20x%2624%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B190%7D%7Bx%7D%3D%5Ccfrac%7B76%7D%7B24%7D%5Cimplies%204560%3D76x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4560%7D%7B76%7D%3Dx%5Cimplies%2060%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btotal%20amount%20of%20students%7D%7D%7B190%2B60%5Cimplies%20250%7D)
Given:
First side(A): 2B - 7
Second side(B): B
Third side(C): B + 4
Plug in values:
A + B + C = 80
2B - 7 + B + B + 4 = 80
If you look at the coefficients only, you can rewrite the equation like this:
2B + B + B - 7 + 4 = 80
This means that 4B - 7 + 4 = 80
=4B - 3 = 80
= 4B = 80 + 3
B = 83/4, so you can either write B as 83/4 or as 20.75.
Checking your work:
A: 2(20.75) - 7 = 34.5
B: 20.75
C: 4 + 20.75 = 24.75
34.5 + 20.75 + 24.75 = 80cm.
Hope this helped.
Answer:
bet
Step-by-step explanation:
Answer:
10 m
Step-by-step explanation:
A square has all equal side lengths. If that is the case, the we have 4 sides that equal each other:
x + x + x + x = 40 m
4x = 40m
x = 10m
So each side is 10 m long.
Answer: 41 more games.
Step-by-step explanation:
4030-2595 to find out how many points he needs to score.
1435
If he averages 35 per game lets divide 1435 by 35 to get the answer of 41.
