1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
Answer:
Left 3, down 2
Step-by-step explanation:
(6,4) --> (3,2)
-3, -2
Left 3, down 2
10 is B
9 is C
11 isB
Hope this helped :)
Answer:
C - b = 190x - 10x
Step-by-step explanation:
Mark brainliest
Answer:
5f + 7
Step-by-step explanation:
8f-3f+2+5
8f-3f is 5f and 2 + 5 is 7
5f+7
