Mark is in charge of the fundraiser for his church. he is selling pre-sale tickets for $10 and at-the-door tickets for $13. the venue has the capacity to hold 275 people. the graph represents the number of raffle tickets he needs to sell to offset the upfront costs and raise at least $3,100:

what is the minimum number of at-the-door tickets he needs to sell to make his goal?
159
158
117
116

Using the data represented by the graph, the minimum number of at-the-door tickets that needs to be sold to meet targeted goal is the value on the y-axis at the point of intersection. Hence, 117 tickets.

The y - axis represents the number of at the door ticket while the x - axis represents the number of pre-sales ticket.

The point of intersection represents the position where the targeted goal is met.

The number of ticket sales at the intersection point is the required value of each ticket sold.

The value on the y-axis which is 116.667 represents, the number of at-the -door tickets.

Therefore, rounding the value to the nearest whole number, then the minimum number of tickets required is 117.

Learn more : link

117

Step-by-step explanation:

At-the-door-tickets cost $13 each.

Pre-sale tickets cost $10 each.

The venue has room for 275 people, so the total number of tickets sold is 275. If all tickets were pre-sale tickets, then he'd make 275 * $10 = $2750. That is less than the amount of money he needs to raise which is $3100.

Since the at-the-door tickets cost $13 each, if he sold all 275 tickets at the door, he'd make 275 * $13 = $3575 which is more than he needs to make.

The question here is what is the minimum number of the more expensive tickets (at $13 each) that he needs to sell and still raise a minimum of $3100.

The intersection of the lines on the graph is at point (158.333, 116.667). The horizontal axis is Pre-sale Tickets, and the vertical axis is At-The-Door Tickets. The y-coordinate representing at-the-door tickets is 116.667. He cannot sell 116.667 tickets. He must sell a whole number of tickets, so he must sell either 116 or 117 tickets.

Let's look at 116 at-the-door tickets:

If he sells 116 at-the-door tickets, then he can sell 275 - 116 = 159 pre-sale tickets.

The money he raises is:

116 * $13 + 159 * $10 = $3098 which is less than $3100

Let's look at 117 at-the-door tickets:

If he sells 117 at-the-door tickets, then he can sell 275 - 117 = 158 pre-sale tickets.

The money he raises is:

117 * $13 + 158 * $10 = $3101 which is more than $3100 and works

A minimum of 117 at-the-door tickets works.

The minimum number of at-the-door tickets is 117.

6.5

step-by-step explanation:

you said it was in the question

~courtney

what's the question here?