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.
For question 2 it is 15
Because: 4 time 6 is 24 and then you minus 9 which is 15
So there you have it the answer is 15
Answer:
Step-by-step explanation:
Y) 
= 3
*multiply both sides by 8 - cancels out 8 in denominator*
x + 4 = 24
*subtract 4 from both sides*
x = 20
E) 
= 1
*multiply both sides by 2 - cancels out 2 in denominator*
x - 5 = 2
*add 5 on both sides*
x = 7
N) 
= 2
* multiply both sides by 4 - cancels out 4 in denominator*
x + 2 = 8
*subtract 2 from both sides*
x = 6
