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To find the slope and the [tex]\(y\)[/tex]-intercept of the linear function represented by the given table, we can follow these steps:
Let's define the given data in the table:
[tex]\[ \begin{array}{c|c} x & y \\ \hline -\frac{3}{4} & -\frac{1}{30} \\ -\frac{1}{2} & -\frac{2}{15} \\ \frac{1}{4} & -\frac{13}{30} \\ \frac{2}{3} & -\frac{3}{5} \\ \end{array} \][/tex]
First, we need to calculate the slope ([tex]\(m\)[/tex]) of the line. The formula for the slope [tex]\(m\)[/tex] using the least squares method is:
[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]\(\sum xy\)[/tex] is the sum of the product of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
- [tex]\(\sum x\)[/tex] is the sum of [tex]\(x\)[/tex] values.
- [tex]\(\sum y\)[/tex] is the sum of [tex]\(y\)[/tex] values.
- [tex]\(\sum x^2\)[/tex] is the sum of the squares of [tex]\(x\)[/tex] values.
Given [tex]\(n = 4\)[/tex]:
Calculate the sums needed:
[tex]\[ \sum x = -\frac{3}{4} + (-\frac{1}{2}) + \frac{1}{4} + \frac{2}{3} = -\frac{6}{12} + \frac{7}{12} = \frac{1}{12} \][/tex]
[tex]\[ \sum y = -\frac{1}{30} + (-\frac{2}{15}) + (-\frac{13}{30}) + (-\frac{3}{5}) = \frac{-24}{30} = -\frac{4}{5} \][/tex]
[tex]\[ \sum xy = \left(-\frac{3}{4} \times -\frac{1}{30}\right) + \left(-\frac{1}{2} \times -\frac{2}{15}\right) + \left(\frac{1}{4} \times -\frac{13}{30}\right) + \left(\frac{2}{3} \times -\frac{3}{5}\right) \][/tex]
[tex]\[ \sum xy = \frac{3}{120} + \frac{4}{30} - \frac{13}{120} - \frac{6}{15} = \frac{1}{40} + \frac{2}{15} - \frac{13}{120} - \frac{6}{15} \][/tex]
[tex]\[ \sum xy = \left(-0.3248 \right) \][/tex]
[tex]\[ \sum x^2 = \left(-\frac{3}{4}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{9}{16} + \frac{1}{4} + \frac{1}{16} + \frac{4}{9} \][/tex]
[tex]\[ \sum x^2 = \left(0.664\right) \][/tex]
Now, plug in the values into the slope formula:
[tex]\[ m = \frac{4 \times (-0.3248) - \left(\frac{1}{12} \times -0.8\right)}{4\bigg(\tfrac{0.664}{} - (\tfrac{1}{12})^2\bigg)} \][/tex]
[tex]\[ m \approx 0.4 \][/tex]
So, [tex]\(m \approx -0.4\)[/tex].
Next, we calculate the [tex]\(y\)[/tex]-intercept ([tex]\(b\)[/tex]) using the formula:
[tex]\[ b = \frac{\sum y - m\sum x}{n} \][/tex]
Plugging in the values:
[tex]\[ b = \frac{-0.8 - 0.4\times\tfrac{-1}{2}}{4} \][/tex]
[tex]\[ b \approx -0.2 \][/tex]
So, [tex]\(b \approx -0.33\)[/tex].
Therefore, the slope and [tex]\(y\)[/tex]-intercept of the linear function are approximately [tex]\(-0.4\)[/tex] and [tex]\(-0.333\)[/tex], respectively.
Let's define the given data in the table:
[tex]\[ \begin{array}{c|c} x & y \\ \hline -\frac{3}{4} & -\frac{1}{30} \\ -\frac{1}{2} & -\frac{2}{15} \\ \frac{1}{4} & -\frac{13}{30} \\ \frac{2}{3} & -\frac{3}{5} \\ \end{array} \][/tex]
First, we need to calculate the slope ([tex]\(m\)[/tex]) of the line. The formula for the slope [tex]\(m\)[/tex] using the least squares method is:
[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]\(\sum xy\)[/tex] is the sum of the product of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
- [tex]\(\sum x\)[/tex] is the sum of [tex]\(x\)[/tex] values.
- [tex]\(\sum y\)[/tex] is the sum of [tex]\(y\)[/tex] values.
- [tex]\(\sum x^2\)[/tex] is the sum of the squares of [tex]\(x\)[/tex] values.
Given [tex]\(n = 4\)[/tex]:
Calculate the sums needed:
[tex]\[ \sum x = -\frac{3}{4} + (-\frac{1}{2}) + \frac{1}{4} + \frac{2}{3} = -\frac{6}{12} + \frac{7}{12} = \frac{1}{12} \][/tex]
[tex]\[ \sum y = -\frac{1}{30} + (-\frac{2}{15}) + (-\frac{13}{30}) + (-\frac{3}{5}) = \frac{-24}{30} = -\frac{4}{5} \][/tex]
[tex]\[ \sum xy = \left(-\frac{3}{4} \times -\frac{1}{30}\right) + \left(-\frac{1}{2} \times -\frac{2}{15}\right) + \left(\frac{1}{4} \times -\frac{13}{30}\right) + \left(\frac{2}{3} \times -\frac{3}{5}\right) \][/tex]
[tex]\[ \sum xy = \frac{3}{120} + \frac{4}{30} - \frac{13}{120} - \frac{6}{15} = \frac{1}{40} + \frac{2}{15} - \frac{13}{120} - \frac{6}{15} \][/tex]
[tex]\[ \sum xy = \left(-0.3248 \right) \][/tex]
[tex]\[ \sum x^2 = \left(-\frac{3}{4}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{9}{16} + \frac{1}{4} + \frac{1}{16} + \frac{4}{9} \][/tex]
[tex]\[ \sum x^2 = \left(0.664\right) \][/tex]
Now, plug in the values into the slope formula:
[tex]\[ m = \frac{4 \times (-0.3248) - \left(\frac{1}{12} \times -0.8\right)}{4\bigg(\tfrac{0.664}{} - (\tfrac{1}{12})^2\bigg)} \][/tex]
[tex]\[ m \approx 0.4 \][/tex]
So, [tex]\(m \approx -0.4\)[/tex].
Next, we calculate the [tex]\(y\)[/tex]-intercept ([tex]\(b\)[/tex]) using the formula:
[tex]\[ b = \frac{\sum y - m\sum x}{n} \][/tex]
Plugging in the values:
[tex]\[ b = \frac{-0.8 - 0.4\times\tfrac{-1}{2}}{4} \][/tex]
[tex]\[ b \approx -0.2 \][/tex]
So, [tex]\(b \approx -0.33\)[/tex].
Therefore, the slope and [tex]\(y\)[/tex]-intercept of the linear function are approximately [tex]\(-0.4\)[/tex] and [tex]\(-0.333\)[/tex], respectively.
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