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?
o y = 0.6x + 0.6
o y = 0.1x + 0.8
o y = 0.8x + 0.1
o y = 0.5x + 0.6

Explanation:

Step1: Recall line - of - best - fit concept

The line of best fit for a set of data points $(x_i,y_i)$ has the form $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. One way to estimate is to use a graphing utility or a statistical software. Another way is to use the fact that we can check by substituting some of the data points into the equations.
Let's take two points, say $(0,2)$ and $(4,5)$.
The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{5 - 2}{4-0}=\frac{3}{4}=0.75$. The y - intercept $b$ is the value of $y$ when $x = 0$. From the point $(0,2)$, $b = 2$ approximately. But we can also test the given equations by substituting $x$ values from the data points and seeing which equation gives the closest $y$ values.
Let's use the point - by - point substitution method.
For the point $(0,2)$:

• For $y=0.6x + 0.6$, when $x = 0$, $y=0.6\times0 + 0.6=0.6$
• For $y=0.1x + 0.8$, when $x = 0$, $y=0.1\times0+0.8 = 0.8$
• For $y=0.8x + 0.1$, when $x = 0$, $y=0.8\times0+0.1=0.1$
• For $y=0.5x + 0.6$, when $x = 0$, $y=0.5\times0+0.6 = 0.6$

Let's use more points. Take the point $(4,5)$:

• For $y=0.6x + 0.6$, when $x = 4$, $y=0.6\times4+0.6=2.4 + 0.6=3$
• For $y=0.1x + 0.8$, when $x = 4$, $y=0.1\times4+0.8=1.2$
• For $y=0.8x + 0.1$, when $x = 4$, $y=0.8\times4+0.1=3.2 + 0.1=3.3$
• For $y=0.5x + 0.6$, when $x = 4$, $y=0.5\times4+0.6=2 + 0.6=2.6$

A more accurate way is to use a statistical formula for the least - squares regression line. The formula for the slope $m$ of the least - squares regression line is $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i=1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and $b=\bar{y}-m\bar{x}$, where $n$ is the number of data points, $\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i$ and $\bar{y}=\frac{1}{n}\sum_{i = 1}^{n}y_i$.
We have $n = 6$, $\sum_{i=1}^{6}x_i=-5 + 1+4-8 + 0+9=1$, $\sum_{i=1}^{6}y_i=0-3 + 5-6 + 2+6=4$, $\sum_{i=1}^{6}x_i^{2}=25 + 1+16 + 64+0 + 81=187$, $\sum_{i=1}^{6}x_iy_i=(-5)\times0+1\times(-3)+4\times5+(-8)\times(-6)+0\times2+9\times6=0-3 + 20 + 48+0 + 54=119$

$m=\frac{6\times119-1\times4}{6\times187 - 1^{2}}=\frac{714 - 4}{1122-1}=\frac{710}{1121}\approx0.63$

$\bar{x}=\frac{1}{6}\approx0.17$, $\bar{y}=\frac{4}{6}\approx0.67$

$b=\bar{y}-m\bar{x}\approx0.67-0.63\times0.17\approx0.67 - 0.11=0.56\approx0.6$

Answer:

$y = 0.6x+0.6$