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:
can't see the picture all the way sorry.
Step-by-step explanation:
Let x = new width
18.6/1.5 = 6.2/x
<span>cross multiply and solve for x
</span>
18.6x = 6.2 * 1.5
18.6x = 9.3
Divide both sides by 18.6
18.6x/18.6 = 9.3/18.6
Simplify
x =0.5
So when height = 6.2, width = 0.5
When trying to cover an area with anything always round up or add a small amount. if you round down for paint or pesticide or anything to cover an area you will need exactly enough or more if dale rounds down he would have .3 square meters uncovered by the pesticide.