Let's determine the slope [tex]\( m \)[/tex] of the linear relation given the points:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 1 & 3 & 6 & 8 & 11 \\
\hline
y & 6.5 & 3.5 & -1 & -4 & -8.5 \\
\hline
\end{array}
\][/tex]
Step 1: Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
The mean of [tex]\( x \)[/tex] ([tex]\( \bar{x} \)[/tex]):
[tex]\[
\bar{x} = \frac{1 + 3 + 6 + 8 + 11}{5} = \frac{29}{5} = 5.8
\][/tex]
The mean of [tex]\( y \)[/tex] ([tex]\( \bar{y} \)[/tex]):
[tex]\[
\bar{y} = \frac{6.5 + 3.5 - 1 - 4 - 8.5}{5} = \frac{-4.5}{5} = -0.7
\][/tex]
Step 2: Calculate the numerator and denominator for the slope
The numerator is:
[tex]\[
\sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})
\][/tex]
The denominator is:
[tex]\[
\sum_{i=1}^{n} (x_i - \bar{x})^2
\][/tex]
Numerator:
[tex]\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = (1-5.8)(6.5+0.7) + (3-5.8)(3.5+0.7) + (6-5.8)(-1+0.7) + (8-5.8)(-4+0.7) + (11-5.8)(-8.5+0.7) = -94.2
\][/tex]
Denominator:
[tex]\[
\sum (x_i - \bar{x})^2 = (1-5.8)^2 + (3-5.8)^2 + (6-5.8)^2 + (8-5.8)^2 + (11-5.8)^2 = 62.8
\][/tex]
Step 3: Calculate the slope [tex]\( m \)[/tex]
[tex]\[
m = \frac{\text{numerator}}{\text{denominator}} = \frac{-94.2}{62.8} = -1.5
\][/tex]
So, the slope [tex]\( m \)[/tex] of the linear relation is [tex]\( -1.5 \)[/tex].
Step 4: Determine the output value when [tex]\( x = 12 \)[/tex]
We use the linear equation [tex]\( y = \bar{y} + m(x - \bar{x}) \)[/tex]:
[tex]\[
y = -0.7 + (-1.5)(12 - 5.8)
\][/tex]
[tex]\[
y = -0.7 + (-1.5) \times 6.2
\][/tex]
[tex]\[
y = -0.7 - 9.3
\][/tex]
[tex]\[
y = -10.0
\][/tex]
Therefore, when [tex]\( x = 12 \)[/tex], the output value will be [tex]\( -10.0 \)[/tex].
