Answer:
a) $169.2
b) $202.8
c) the church can bring either 31 or 34 members
Step-by-step explanation:
For each person the church brings that exceeds 15, the ticket price goes down by $0.20
18 members exceed 15 by 3.
This means that for 3 persons the church brings that exceeds 15, we have 3*0.20 = $0.60
The price of the ticket will now be
10 - 0.6 = $9.40
For the 18 members, we now have
18*9.4 = $169.2
b) The model for bringing a number of people at a particular cost = C(x)
x is the number of people that exceeds 15
C(x) = (15+x)(10-0.2x)
C(x) = 150 - 3x + 10x - 0.2x^2
= 150 + 7x - 0.2x^2
In bringing 26 people to the show, it is 11 more than 15
C(11) = 150 + 7(11) - 0.2(11)^2
C(11) = 150 +77 - 24.2
= $202.8
c) if the church can only afford $210.8, to solve x, we have
150 + 7x - 0.2x^2 = 210.8
-0.2x^2 + 7x + 150 = 210.8
-0.2x^2 + 7x + 150 - 210.8 = 0
-0.2x^2 + 7x -60.8 = 0
(-b +/-√b^2 - 4ac) / 2a
a = -0.2 , b = 7 c = -60.8
(-7 +/-√7^2 - 4(-0.2)(-60.8)) / 2(-0.2) = 0
(-7 +/-√49 - 48.64) / -0.4 = 0
(-7 +/-√0.36) / -0.4 = 0
(-7 +/- 0.6) / -0.4 = 0
(-7+0.6)/-0.4 or (-7 - 0.6)/-0.4
(-6.4)/-0.4 or (-7.6)/- 0.4
16 or 19
So we have (15+16) = 31 members
(15+19) = 34 members
The church can bring either 31 members or 34 members
