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Question 1
An SAT tutoring services charges an initial fee of $500 plus $50 for each additional session. David wants to spend $700 at most on tutoring.
Which inequality should be used to determine the number of additional sessions (s) David can schedule?
(1 point)
Responses
50s +500>700
50s +500>700
50s + 500<700
50s + 500<700
50s + 500 ≤700
50s + 500 ≤700
50s + 500 ≥700
50s + 500 ≥700
Question 2
(1 point)
David can schedule
additional sessions.
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Answers
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Answered by
GPT-4o mini
To determine the number of additional sessions (s) that David can schedule given that he wants to spend at most $700 on tutoring, we need to set up an inequality based on the cost structure.
The cost David incurs for the tutoring sessions is comprised of:
- An initial fee of $500
- An additional fee of $50 for each session he schedules
Thus, the total cost can be written as:
\[ 500 + 50s \]
Since David wants to spend at most $700, we can set up the inequality:
\[ 500 + 50s \leq 700 \]
Now let's review the given options:
1. \( 50s + 500 > 700 \)
2. \( 50s + 500 < 700 \)
3. \( 50s + 500 \leq 700 \) (correct)
4. \( 50s + 500 \geq 700 \)
The correct choice is:
**50s + 500 ≤ 700**
Now, we can find out how many additional sessions David can schedule.
To solve for s:
\[ 500 + 50s \leq 700 \]
Subtract 500 from both sides:
\[ 50s \leq 200 \]
Now, divide by 50:
\[ s \leq 4 \]
Thus, David can schedule **4 additional sessions**.
The cost David incurs for the tutoring sessions is comprised of:
- An initial fee of $500
- An additional fee of $50 for each session he schedules
Thus, the total cost can be written as:
\[ 500 + 50s \]
Since David wants to spend at most $700, we can set up the inequality:
\[ 500 + 50s \leq 700 \]
Now let's review the given options:
1. \( 50s + 500 > 700 \)
2. \( 50s + 500 < 700 \)
3. \( 50s + 500 \leq 700 \) (correct)
4. \( 50s + 500 \geq 700 \)
The correct choice is:
**50s + 500 ≤ 700**
Now, we can find out how many additional sessions David can schedule.
To solve for s:
\[ 500 + 50s \leq 700 \]
Subtract 500 from both sides:
\[ 50s \leq 200 \]
Now, divide by 50:
\[ s \leq 4 \]
Thus, David can schedule **4 additional sessions**.
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