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Question 1
Kelly is attending a fall carnival with only $10. Admission into the carnival is $3 and each game inside the carnival costs $0.25.

Which inequality can be used to determine the maximum number games Kelly can play?

(1 point)
Responses

0.25x+3≤10
0 point 2 5 x plus 3 is less than or equal to 10

0.25x+3≥10
0 point 2 5 x plus 3 is greater than or equal to 10

0.25x+3<10
0.25x+3<10

0.25x+3>10
0 point 2 5 x plus 3 is greater than 10
Question 2
(1 point)
Kelly has enough money to play at most
games.
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1 answer

To solve the problem, let's break it down step by step.

  1. Admission cost: Kelly needs to pay $3 to enter the carnival.
  2. Amount left for games: After paying for admission, Kelly has $10 - $3 = $7 left.
  3. Cost per game: Each game costs $0.25.

Let \( x \) be the number of games Kelly can play. The total cost for \( x \) games would be \( 0.25x \). To find the maximum number of games she can play, we set up the following inequality:

\[ 3 + 0.25x \leq 10 \]

This simplifies to:

\[ 0.25x \leq 7 \]

Now we can determine which of the provided inequalities is correct:

  • Option 1: \( 0.25x + 3 \leq 10 \) is the correct inequality because it represents the total money spent being less than or equal to her total amount ($10).
  • Option 2: \( 0.25x + 3 \geq 10 \) is incorrect.
  • Option 3: \( 0.25x + 3 < 10 \) is also technically correct but does not represent the maximum number of games because it would allow for games only below the maximum.
  • Option 4: \( 0.25x + 3 > 10 \) is incorrect.

Thus, the correct inequality is:

0.25x + 3 ≤ 10


Next, let's compute the maximum number of games Kelly can play:

From the inequality \( 0.25x + 3 \leq 10 \):

  1. Subtract 3 from both sides: \[ 0.25x \leq 7 \]

  2. Divide both sides by 0.25: \[ x \leq \frac{7}{0.25} = 28 \]

So, Kelly can play at most 28 games.

To summarize:

  • Kelly can play at most 28 games.