Question

raquel throws darts at a coordinate grid centered at the origin. her goal is to create a line of darts. her darts actually hit the coordinate grid at (-5, 0), (1, -3), (4, 5), (-8, -6), (0, 2), and (9, 6). which equation best approximates the line of best fit of the darts?
( y = 0.6x + 0.6 )
( y = 0.1x + 0.8 )
( y = 0.8x + 0.1 )
( y = 0.5x + 0.6 )

Explanation:

Step1: Calculate the mean of \(x\) - values and \(y\) - values

The \(x\) - values are \(x_1=-5,x_2 = 1,x_3=4,x_4=-8,x_5 = 0,x_6=9\).
The mean of \(x\) - values \(\bar{x}=\frac{-5 + 1+4-8 + 0+9}{6}=\frac{1}{6}\approx0.17\)
The \(y\) - values are \(y_1 = 0,y_2=-3,y_3=5,y_4=-6,y_5=2,y_6=6\)
The mean of \(y\) - values \(\bar{y}=\frac{0-3 + 5-6+2+6}{6}=\frac{4}{6}\approx0.67\)

Step2: Calculate the slope \(b_1\)

The formula for the slope \(b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\)
\((x_1-\bar{x})(y_1-\bar{y})=(-5 - 0.17)(0 - 0.67)=(-5.17)(- 0.67)\approx3.46\)
\((x_2-\bar{x})(y_2-\bar{y})=(1 - 0.17)(-3 - 0.67)=(0.83)(-3.67)\approx - 3.05\)
\((x_3-\bar{x})(y_3-\bar{y})=(4 - 0.17)(5 - 0.67)=(3.83)(4.33)\approx16.68\)
\((x_4-\bar{x})(y_4-\bar{y})=(-8 - 0.17)(-6 - 0.67)=(-8.17)(-6.67)\approx54.5\)
\((x_5-\bar{x})(y_5-\bar{y})=(0 - 0.17)(2 - 0.67)=(-0.17)(1.33)\approx - 0.23\)
\((x_6-\bar{x})(y_6-\bar{y})=(9 - 0.17)(6 - 0.67)=(8.83)(5.33)\approx47.07\)
\(\sum_{i = 1}^{6}(x_i-\bar{x})(y_i-\bar{y})\approx3.46-3.05 + 16.68+54.5-0.23+47.07=118.43\)

\((x_1-\bar{x})^2=(-5 - 0.17)^2=(-5.17)^2\approx26.73\)
\((x_2-\bar{x})^2=(1 - 0.17)^2=(0.83)^2\approx0.69\)
\((x_3-\bar{x})^2=(4 - 0.17)^2=(3.83)^2\approx14.67\)
\((x_4-\bar{x})^2=(-8 - 0.17)^2=(-8.17)^2\approx66.75\)
\((x_5-\bar{x})^2=(0 - 0.17)^2=(-0.17)^2\approx0.03\)
\((x_6-\bar{x})^2=(9 - 0.17)^2=(8.83)^2\approx77.97\)
\(\sum_{i = 1}^{6}(x_i-\bar{x})^2\approx26.73+0.69+14.67+66.75+0.03+77.97=186.84\)
\(b_1=\frac{118.43}{186.84}\approx0.63\approx0.6\)

Step3: Calculate the intercept \(b_0\)

Using the formula \(b_0=\bar{y}-b_1\bar{x}\), substituting \(\bar{x}\approx0.17\), \(\bar{y}\approx0.67\) and \(b_1\approx0.6\)
\(b_0=0.67-0.6\times0.17=0.67 - 0.102 = 0.568\approx0.6\)

Answer:

\(y = 0.6x+0.6\)