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.
B= 30
a2+b2=c2
72^2+b^2 = 78^2
5184+b^2 = 6084
b2 = 900
b = 30
A^2+b^2=c^2
5^2+b^2=12^2
25+b^2=144
b^2=144-25
b^2=119
SQUARE ROOT BOTH SIDES:
b=10.9
The answer is c.
(1 + x) (13) = 13+6.5
13x + 13 = 19.5
13x = 6.5
x = 0.5
The item was marked up by 50%
Since 1/4 is equal to 0.25 than you would do 5,400 times 0.25 and get the answer of 13,500.0. So the race car weighs 13,500 pounds.
