chelsea has $45 to spend at the fair. she spends $20. on admissib and $15 on snacks. she wants to play a game that costs $0.65 per game. Write an inequality to find the maximum number of times, x, Chelsa can play games. Using this inequality, determine the maximum number of times she can play the game.

20 + 15 + 0.65x < = 45
35 + 0.65x < = 45
0.65x < = 45 - 35
0.65x < = 10
x < = 10 / 0.65
x < = 15.3......so the greatest number of games she can play is 15

Answer Link

ColinJacobus ColinJacobus · Answer 2 · 2019-08-14T09:53:09+02:00

Answer: The required inequality is [tex]35+0.65x\leq 45[/tex] and teh maximum number of times that Chelsea can play the game is 15.

Step-by-step explanation: Given that Chelsea had $45 to spend at the fair. she spends $20 on admissible and $15 on snacks. She wants to play a game that costs $0.65 per game.

We are to write an inequality to find the maximum number of times, x, Chelsea can play games.

Also, to find the maximum number of times she can play he game.

According to the given information, the inequality can be written as follows :

[tex]20+15+0.65\times x\leq 45\\\\\Rightarrow 35+0.65x\leq 45.[/tex]

And, the solution of the above inequality is as follows :

[tex]35+0.65x\leq 45\\\\\Rightarrow 0.65x\leq 45-35\\\\\Rightarrow 0.65x\leq 10\\\\\Rightarrow x\leq \dfrac{10}{0.65}\\\\\Rightarrow x\leq 15.38.[/tex]

Thus, the maximum number of times Chelsea can play the game is 15.