Each student who buys a school lunch 
pays $2.75. The cafeteria typically 
brings in between $1,168.75 and 
$1,438.25 each day.
a. Model and solve this situation using 
an inequality.
b. Describe in words the quantities 
that would work in this situation.

### Solution:

Let's solve the problem step-by-step by addressing each part.

#### Part (a): Model and Solve the Situation Using an Inequality

1. **Define the Variable:**
   - Let $ x $ represent the number of students buying school lunch each day.

2. **Set Up the Inequality:**
   - Each student pays $2.75 for lunch, so the total revenue for the cafeteria each day can be represented as $ 2.75x $.
   - The cafeteria brings in between $1,168.75 and $1,438.25 each day. Therefore, the inequality that models the situation is:
   \[
   1168.75 \leq 2.75x \leq 1438.25
   \]

3. **Solve the Inequality:**
   - We will solve this compound inequality by dividing all parts of the inequality by 2.75:

   \[
   \frac{1168.75}{2.75} \leq x \leq \frac{1438.25}{2.75}
   \]

   Let's calculate the values:

   \[
   x \geq \frac{1168.75}{2.75} \quad \text{and} \quad x \leq \frac{1438.25}{2.75}
   \]

   Calculating these values:

   \[
   x \geq 425 \quad \text{and} \quad x \leq 523
   \]

   Thus, the inequality for the number of students buying lunch is:

   \[
   425 \leq x \leq 523
   \]

#### Part (b): Describe the Quantities That Would Work in This Situation

In words, the number of students buying a school lunch each day must be between **425 and 523** (inclusive). This means:

- **At least 425 students** need to buy lunch for the cafeteria to make at least $1,168.75.
- **At most 523 students** can buy lunch for the cafeteria to not exceed $1,438.25 in daily revenue.

Any number of students between 425 and 523 would result in the cafeteria's revenue falling within the given range of $1,168.75 to $1,438.25.

Would you like more details on this topic or have any questions?

### Related Questions:
1. How would the inequality change if the lunch price were different?
2. What if the revenue range changed to a different set of values?
3. How would you graph this inequality on a number line?
4. What is the impact on the number of students if the cafeteria wants to raise at least $1,500?
5. How does the problem change if the price is discounted for certain students?

### Tip:
When solving inequalities involving money or real-life quantities, always carefully consider the units and ensure that your inequality covers all possible scenarios.

Each student who buys a school lunch pays $2.75. The cafeteria typically brings in between $1,168.75 and $1,438.25 each day. a. Model and solve this situation using an inequality. b. Describe in words the quantities that would work in this situation.

Model and solve a real-world problem involving school lunch revenue using linear inequalities. Learn how to calculate the number of students needed to generate revenue between $1,168.75 and $1,438.25. Ideal for Grades 6-8, this example explains key algebraic concepts and provides a detailed step-by-step solution.

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