Anna is in charge of the alumni fundraiser for her alma mater. she is selling pre-sale tickets for $10 and at-the-door tickets $25. the venue has the capacity to hold 400 people. the graph represents the number of tickets anna needs to sell to offset her upfront costs and raise at least $5,000 for her school:
what is the minimum number of at-the-door tickets she needs to sell to make her goal?
a,333
b.334
c.66
d.67

D. 67

Step-by-step explanation:

To help me see the region we needed to look at, I wrote the inequalities out.

Let x be number of pre-sale tickets and y be the number of at-the-door tickets as your graph suggests.

So one of the inequalities about number of where the other one is about cost.

You are given x+y is no more than 400 or x+y<=400 (the top line graphed in your picture is x+y=400).

You are given 10x+25y is at least 5000 or 10x+25y>=5000 (the bottom line graphed in your picture).

I solved both of these for y.

x+y<=400

Subtract x on both sides giving y<=-x+400 (shaded below line because of the y< part).

10x+25y>=5000

Subtract 10x on both sides:

       25y>=-10x+5000

Divide both sides by 25:

          y>=-2/5 x+200 (shaded above the line because of y> part).

The region we should then be looking at is:

Let's look at the points (0,400), (0,200), and finally (333,67).

Cost=10x+25y

Let's plug in

Cost=10(0)+25(400)=10000

Cost=10(0)+25(200)=5000  (can we go lower than 200)

Cost=10(333)+25(67)=5005 (our y is lower 200 so far this is the winner)

Cost 10(334)+25(66)=4990 (didn't meet the 5000 dollar requirement)

67 D.

(Also if you look at the graph 66 would not be included in the shaded region; it would be too low)

boxplot

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: