I think this is a least common multiple question
So they would all arrive on day 504, (which probably is a few years)
8 = 2 × 2 × 2
18 = 2 × 3 × 3
21 = 3 × 7
I might be right I might be wrong, IM not positively sure but yknow
154
Subtract $1060 from $5680 to get $4620 > divide $4620 by $30 to get 154 > 154 students can attend
Answer:
Yes
Step-by-step explanation:
A negative multiplied by another negative will turn into a positive. A positive multipled by a negative will stay as a negative.
1.
If no changes are made, the school has a revenue of :
625*400$/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.