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Let's break down this problem step-by-step.

### Given Information:
- In June, Jane used 480 minutes and the cost was $140.
- In July, she used 960 minutes and the cost was $260.

This is a linear problem, where we assume Jane's phone bill is structured as follows:
- There is a fixed monthly fee (base cost).
- There is an additional cost per minute of calling time.

We need to:
1. Find the equation for her monthly cost $ C $ as a function of minutes $ x $.
2. Calculate her bill if she used 744 minutes.

### Step A: Find the linear equation

We know the cost structure fits the form of a linear equation:
\[
C = m \cdot x + b
\]
Where:
- $ C $ is the total cost,
- $ x $ is the number of minutes used,
- $ m $ is the cost per minute,
- $ b $ is the fixed monthly fee.

We have two points based on the given dataLet's break down this problem step-by-step.

### Given Information:
- In June, Jane used 480 minutes and the cost was $140.
- In July, she used 960 minutes and the cost was $260.

This is a linear problem, where we assume Jane's phone bill is structured as follows:
- There is a fixed monthly fee (base cost).
- There is an additional cost per minute of calling time.

We need to:
1. Find the equation for her monthly cost $ C $ as a function of minutes $ x $.
2. Calculate her bill if she used 744 minutes.

### Step A: Find the linear equation

We know the cost structure fits the form of a linear equation:
\[
C = m \cdot x + b
\]
Where:
- $ C $ is the total cost,
- $ x $ is the number of minutes used,
- $ m $ is the cost per minute,
- $ b $ is the fixed monthly fee.

We have two points based on the given 
- (480, 140) — for June
- (960, 260) — for July

We can use these two points to find the slope $ m $ (the cost per minute):
\[
m = \frac{C_2 - C_1}{x_2 - x_1} = \frac{260 - 140}{960 - 480} = \frac{120}{480} = 0.25
\]
So, the cost per minute is $0.25.

Now we can find the fixed monthly fee $ b $ by substituting one of the points into the equation. Let's use the June data point (480, 140):
\[
140 = 0.25 \times 480 + b
\]
\[
140 = 120 + b
\]
\[
b = 140 - 120 = 20
\]

Thus, the equation for the monthly cost is:
\[
C = 0.25x + 20
\]

### Step B: Calculate the bill for 744 minutes

Now that we have the equation $ C = 0.25x + 20 $, we can substitute $ x = 744 $ to find the bill:
\[
C = 0.25 \times 744 + 20
\]
\[
C = 186 + 20 = 206
\]

Thus, her bill for 744 minutes is **$206**.

---

Would you like further details on any part of this solution? Here are some related questions:

1. How would the equation change if the fixed fee increased by $5?
2. What if the cost per minute doubled—how would that affect her bill for 744 minutes?
3. Can you find the break-even point where the cost per minute matches the fixed fee?
4. How would Jane's bill look if she used zero minutes in a month?
5. What happens to the bill if there’s a limit on minutes before extra charges kick in?

**Tip**: For any linear modeling problem, always identify the key variables and look for two points to define the slope.

Jane has a pre-paid cell phone with Splint. She can't remember the exact costs, but her plan has a monthly fee and a charge for each minute of calling time. In June she used 480 minutes and the cost was $140. In July she used 960 minutes and the cost was $260.
A) Express the monthly cost C as a function of x, the number of minutes of calling time she used.
B) If Jane used 744 minutes of calling time in August, how much was her bill?

Learn how to model a phone bill using linear equations based on usage minutes. This problem calculates Jane's monthly cost using 480 and 960 minutes as reference points. Suitable for high school students studying algebra and linear functions.

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