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.
The answer I got was 2c-48 still trying to figure out the matching part
Answer:
The page numbers are 36 and 37.
Step-by-step explanation:
n + (n + 1) = 73
2n + 1 = 73
n = (73 - 1)/2
= 72/2 = 36
Answer:
x^3 + 1
Step-by-step explanation:
Swap x and y and solve for y:
x =(y-1)^⅓
x³ = y-1
y = x³+1
f(x) = x³+1