Question

Pete wants to treat several friends to the baseball game. He has a budget of $110. Tickets are $8. He will spend $20 on parking. How many friends can he afford to take without going over his budget?

I also need a graph, a solution, and an explanation, plsssssssss help asap!

100 points!

Answer 1

11 friends

Explanation:

Let's call the number of friends that Pete can afford to take "x".

We know that the cost of the tickets and parking for x friends cannot exceed Pete's budget of $110:

8x + 20 ≤ 110

Subtracting 20 from both sides of the inequality, we get:

8x ≤ 90

Dividing both sides by 8, we get:

x ≤ 11.25

Since Pete cannot take a fractional number of friends, he must round down to the nearest whole number. Therefore, he can afford to take 11 friends to the game without going over his budget.

Here's your graph:

Answer 2

11

Explanation:

$110 budget-$20 parking=$90

8×12 =96 so do 8x11=88