Answered

Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

The student Fun Club plans to go to the movies. At the matinee, tickets cost $6 and popcorn is $3. At evening shows, tickets cost $9 and popcorn is $4. The Fun Club attends a matinee and spends less than $60, and then attends an evening show and spends more than $36. If they purchased the same number of tickets and popcorns at each show, which of the following is a possible solution for the number of tickets and popcorns purchased?

The Student Fun Club Plans To Go To The Movies At The Matinee Tickets Cost 6 And Popcorn Is 3 At Evening Shows Tickets Cost 9 And Popcorn Is 4 The Fun Club Atte class=

Sagot :

Matinee

Cost of each ticket: $6

Cost of popcorn: $3

Evening:

Ticket: $9

Popcorn: $4

Number of tickets: x

Number of popcorns : y

The Fun Club attends a matinee and spends less than $60

6x + 3y < 60

Then attends an evening show and spends more than $36

9x+ 4y < 36

We have the system:

6x + 3y < 60 (a)

9x+ 4y >36 (b)

Graph each inequality:

The intersection of red and blue is the solution.

7 tickets and 5 popcorns (7,5) is inside the intersection, So, it is the solution.

View image ElaX97654
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.