Question
The linear equation represented on the coordinate plane represents the average cost of a loaf of white bread since the year 1985.
(10, 0.85) (17, 1.06) (23, 1.24) (30, 1.45)
Select all the statements that are true for the linear equation.
(4 points)
Responses
y = 0.03x + 0.55 represents the linear equation
y = 0.03x + 0.55 represents the linear equation
The cost of a loaf of bread in 1990 was $0.70.
The cost of a loaf of bread in 1990 was $0.70.
(27,1.48) is a solution to the linear equation
(27,1.48) is a solution to the linear equation
The cost of a loaf of bread in 2015 was $1.50
The cost of a loaf of bread in 2015 was $1.50
(5, 0.7) is a solution to the linear equation
(5, 0.7) is a solution to the linear equation
A loaf of bread cost $0.55 in 1985
(10, 0.85) (17, 1.06) (23, 1.24) (30, 1.45)
Select all the statements that are true for the linear equation.
(4 points)
Responses
y = 0.03x + 0.55 represents the linear equation
y = 0.03x + 0.55 represents the linear equation
The cost of a loaf of bread in 1990 was $0.70.
The cost of a loaf of bread in 1990 was $0.70.
(27,1.48) is a solution to the linear equation
(27,1.48) is a solution to the linear equation
The cost of a loaf of bread in 2015 was $1.50
The cost of a loaf of bread in 2015 was $1.50
(5, 0.7) is a solution to the linear equation
(5, 0.7) is a solution to the linear equation
A loaf of bread cost $0.55 in 1985
Answers
GPT-4o mini
To determine which statements are true regarding the linear equation representing the average cost of a loaf of white bread since 1985, we need to analyze the given coordinates and the proposed linear equation.
1. **Finding the linear equation:**
Given points are:
- (10, 0.85), which represents the year 1995 (10 years after 1985)
- (17, 1.06), which represents the year 2002 (17 years after 1985)
- (23, 1.24), which represents the year 2008 (23 years after 1985)
- (30, 1.45), which represents the year 2015 (30 years after 1985)
To find the slope (m) of the line connecting any two points, we can use the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Let's take the points (10, 0.85) and (30, 1.45):
\[
m = \frac{1.45 - 0.85}{30 - 10} = \frac{0.60}{20} = 0.03
\]
Now we can find the y-intercept (b) using the equation of a line \(y = mx + b\). Using the point (10, 0.85):
\[
0.85 = 0.03(10) + b \implies 0.85 = 0.30 + b \implies b = 0.55
\]
Therefore, the equation of the line is:
\[
y = 0.03x + 0.55
\]
2. **Evaluating each statement:**
- **y = 0.03x + 0.55 represents the linear equation.**
- **True** (confirmed by our calculation)
- **The cost of a loaf of bread in 1990 was $0.70.**
- 1990 corresponds to \(x = 5\):
\[
y = 0.03(5) + 0.55 = 0.15 + 0.55 = 0.70
\]
- **True**
- **(27, 1.48) is a solution to the linear equation.**
- Plugging \(x = 27\):
\[
y = 0.03(27) + 0.55 = 0.81 + 0.55 = 1.36
\]
- Since \(1.48 \neq 1.36\), this statement is **False**.
- **The cost of a loaf of bread in 2015 was $1.50.**
- 2015 corresponds to \(x = 30\):
\[
y = 0.03(30) + 0.55 = 0.90 + 0.55 = 1.45
\]
- Since $1.50 does not equal $1.45, this statement is **False**.
- **(5, 0.7) is a solution to the linear equation.**
- Plugging \(x = 5\):
\[
y = 0.03(5) + 0.55 = 0.70
\]
- **True**
- **A loaf of bread cost $0.55 in 1985.**
- The year 1985 corresponds to \(x = 0\):
\[
y = 0.03(0) + 0.55 = 0.55
\]
- **True**
**In conclusion, the true statements are:**
1. y = 0.03x + 0.55 represents the linear equation.
2. The cost of a loaf of bread in 1990 was $0.70.
3. (5, 0.7) is a solution to the linear equation.
4. A loaf of bread cost $0.55 in 1985.
1. **Finding the linear equation:**
Given points are:
- (10, 0.85), which represents the year 1995 (10 years after 1985)
- (17, 1.06), which represents the year 2002 (17 years after 1985)
- (23, 1.24), which represents the year 2008 (23 years after 1985)
- (30, 1.45), which represents the year 2015 (30 years after 1985)
To find the slope (m) of the line connecting any two points, we can use the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Let's take the points (10, 0.85) and (30, 1.45):
\[
m = \frac{1.45 - 0.85}{30 - 10} = \frac{0.60}{20} = 0.03
\]
Now we can find the y-intercept (b) using the equation of a line \(y = mx + b\). Using the point (10, 0.85):
\[
0.85 = 0.03(10) + b \implies 0.85 = 0.30 + b \implies b = 0.55
\]
Therefore, the equation of the line is:
\[
y = 0.03x + 0.55
\]
2. **Evaluating each statement:**
- **y = 0.03x + 0.55 represents the linear equation.**
- **True** (confirmed by our calculation)
- **The cost of a loaf of bread in 1990 was $0.70.**
- 1990 corresponds to \(x = 5\):
\[
y = 0.03(5) + 0.55 = 0.15 + 0.55 = 0.70
\]
- **True**
- **(27, 1.48) is a solution to the linear equation.**
- Plugging \(x = 27\):
\[
y = 0.03(27) + 0.55 = 0.81 + 0.55 = 1.36
\]
- Since \(1.48 \neq 1.36\), this statement is **False**.
- **The cost of a loaf of bread in 2015 was $1.50.**
- 2015 corresponds to \(x = 30\):
\[
y = 0.03(30) + 0.55 = 0.90 + 0.55 = 1.45
\]
- Since $1.50 does not equal $1.45, this statement is **False**.
- **(5, 0.7) is a solution to the linear equation.**
- Plugging \(x = 5\):
\[
y = 0.03(5) + 0.55 = 0.70
\]
- **True**
- **A loaf of bread cost $0.55 in 1985.**
- The year 1985 corresponds to \(x = 0\):
\[
y = 0.03(0) + 0.55 = 0.55
\]
- **True**
**In conclusion, the true statements are:**
1. y = 0.03x + 0.55 represents the linear equation.
2. The cost of a loaf of bread in 1990 was $0.70.
3. (5, 0.7) is a solution to the linear equation.
4. A loaf of bread cost $0.55 in 1985.