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Sagot :
To solve this problem meticulously, let’s start by explaining each step of the solution:
### Step 1: Understanding the Problem
We need to plot the principal balance over time as a graph where the month is the independent variable and the principal balance is the dependent variable. We aim to observe the relationship and find a linear best fit function for this data.
### Step 2: Extract the Data Points
We extract the monthly data from the table. Let's organize it into two arrays: one for the months and one for the principal balances.
- Months (Independent Variable): [tex]\( x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] \)[/tex]
- Principal Balance (Dependent Variable): [tex]\( y = [16791.47, 16331.80, 15870.98, 15409.01, 14945.88, 14481.59, 14016.15, 13549.54, 13061.76, 12612.81, 12142.69, 11671.40] \)[/tex]
### Step 3: Plotting the Data Points
Imagine plotting these data points on a scatter plot with the months on the x-axis and the principal balances on the y-axis.
### Step 4: Finding the Linear Best Fit
To find the linear best fit to these points, we use the method of least squares to calculate the slope ([tex]\(m\)[/tex]) and y-intercept ([tex]\(b\)[/tex]) of the best fit line. The linear best fit can be written in the form:
[tex]\[ y = mx + b \][/tex]
### Step 5: Calculation
#### Compute the slope ([tex]\(m\)[/tex]):
The slope of the line [tex]\( m \)[/tex] is calculated using the formula:
[tex]\[ m = \frac{N \sum{xy} - \sum{x} \sum{y}}{N \sum{x^2} - (\sum{x})^2} \][/tex]
Where [tex]\( N \)[/tex] is the number of data points, [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are arrays of the independent and dependent variables, respectively. Here, we have 12 data points (N = 12).
#### Compute the y-intercept ([tex]\(b\)[/tex]):
The y-intercept [tex]\( b \)[/tex] is calculated using the formula:
[tex]\[ b = \frac{\sum{y} - m \sum{x}}{N} \][/tex]
### Step 6: Perform the Calculations:
First, calculate the necessary sums:
[tex]\[ \sum{x} = 1 + 2 + \ldots + 12 = 78 \][/tex]
[tex]\[ \sum{y} = 16791.47 + 16331.80 + \ldots + 11671.40 = 167386.08 \][/tex]
[tex]\[ \sum{xy} = (1 \times 16791.47) + (2 \times 16331.80) + \ldots + (12 \times 11671.40) = 1040203.74 \][/tex]
[tex]\[ \sum{x^2} = 1^2 + 2^2 + \ldots + 12^2 = 650 \][/tex]
Substitute these sums into the formulas for [tex]\( m \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ m = \frac{12 \times 1040203.74 - 78 \times 167386.08}{12 \times 650 - 78^2} \approx \frac{12482444.88 - 13056234.24}{7800 - 6084} \approx \frac{-573789.36}{1716} \approx -334.44 \][/tex]
[tex]\[ b = \frac{167386.08 - (-334.44 \times 78)}{12} \approx \frac{167386.08 + 26086.32}{12} \approx \frac{193472.40}{12} \approx 16122.70 \][/tex]
### Step 7: Conclusion
The best fit linear function that describes the relationship between the month and the principal balance is:
[tex]\[ y = -334.44x + 16122.70 \][/tex]
This means that for every additional month, the principal balance decreases by approximately [tex]$334.44, starting from an approximate initial principal balance of \$[/tex]16122.70 in the first month.
### Step 1: Understanding the Problem
We need to plot the principal balance over time as a graph where the month is the independent variable and the principal balance is the dependent variable. We aim to observe the relationship and find a linear best fit function for this data.
### Step 2: Extract the Data Points
We extract the monthly data from the table. Let's organize it into two arrays: one for the months and one for the principal balances.
- Months (Independent Variable): [tex]\( x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] \)[/tex]
- Principal Balance (Dependent Variable): [tex]\( y = [16791.47, 16331.80, 15870.98, 15409.01, 14945.88, 14481.59, 14016.15, 13549.54, 13061.76, 12612.81, 12142.69, 11671.40] \)[/tex]
### Step 3: Plotting the Data Points
Imagine plotting these data points on a scatter plot with the months on the x-axis and the principal balances on the y-axis.
### Step 4: Finding the Linear Best Fit
To find the linear best fit to these points, we use the method of least squares to calculate the slope ([tex]\(m\)[/tex]) and y-intercept ([tex]\(b\)[/tex]) of the best fit line. The linear best fit can be written in the form:
[tex]\[ y = mx + b \][/tex]
### Step 5: Calculation
#### Compute the slope ([tex]\(m\)[/tex]):
The slope of the line [tex]\( m \)[/tex] is calculated using the formula:
[tex]\[ m = \frac{N \sum{xy} - \sum{x} \sum{y}}{N \sum{x^2} - (\sum{x})^2} \][/tex]
Where [tex]\( N \)[/tex] is the number of data points, [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are arrays of the independent and dependent variables, respectively. Here, we have 12 data points (N = 12).
#### Compute the y-intercept ([tex]\(b\)[/tex]):
The y-intercept [tex]\( b \)[/tex] is calculated using the formula:
[tex]\[ b = \frac{\sum{y} - m \sum{x}}{N} \][/tex]
### Step 6: Perform the Calculations:
First, calculate the necessary sums:
[tex]\[ \sum{x} = 1 + 2 + \ldots + 12 = 78 \][/tex]
[tex]\[ \sum{y} = 16791.47 + 16331.80 + \ldots + 11671.40 = 167386.08 \][/tex]
[tex]\[ \sum{xy} = (1 \times 16791.47) + (2 \times 16331.80) + \ldots + (12 \times 11671.40) = 1040203.74 \][/tex]
[tex]\[ \sum{x^2} = 1^2 + 2^2 + \ldots + 12^2 = 650 \][/tex]
Substitute these sums into the formulas for [tex]\( m \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ m = \frac{12 \times 1040203.74 - 78 \times 167386.08}{12 \times 650 - 78^2} \approx \frac{12482444.88 - 13056234.24}{7800 - 6084} \approx \frac{-573789.36}{1716} \approx -334.44 \][/tex]
[tex]\[ b = \frac{167386.08 - (-334.44 \times 78)}{12} \approx \frac{167386.08 + 26086.32}{12} \approx \frac{193472.40}{12} \approx 16122.70 \][/tex]
### Step 7: Conclusion
The best fit linear function that describes the relationship between the month and the principal balance is:
[tex]\[ y = -334.44x + 16122.70 \][/tex]
This means that for every additional month, the principal balance decreases by approximately [tex]$334.44, starting from an approximate initial principal balance of \$[/tex]16122.70 in the first month.
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