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:
1.5 for half a batch and 1 and 0.16666666.. batch
Step-by-step explanation:
3 cups=1 batch so 3/2= 1.5 cups=1/2 batch
4 cups=1 and .1666 batches
Answer:
180°
because x is the mid pf 0° and 360°
Answer: 4
Step-by-step explanation: 7+-3 = 4
Answer:
Step-by-step explanation:
(9+(15*15))+7=241
