Question

darius is studying the relationship between mathematics and art. he asks friends to each draw a \typical\ rectangle. he measures the length and width in centimeters of each rectangle and plots the points on a graph, where x represents the width and y represents the length. the points representing the rectangles are (6.1, 12.0), (5.0, 8.1), (9.1, 15.2), (6.5, 10.2), (7.4, 11.3), and (10.9, 17.5). which equation could darius use to determine the length, in centimeters, of a \typical\ rectangle for a given width in centimeters?
o y = 0.605x + 0.004
o y = 0.959x + 0.041
o y = 1.518x + 0.995
o y = 1.967x + 0.984

Explanation:

Step1: Recall the form of a linear - regression equation

The general form of a linear - regression equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. We can use a statistical method (such as the least - squares method) to find the best - fit line for the given data points $(x_1,y_1),(x_2,y_2),\cdots,(x_n,y_n)$. In this case, we can also check each equation by substituting the $x$ - values of the data points into the equations and seeing how well the resulting $y$ - values match the actual $y$ - values.
Let's take the first data point $(6.1,12.0)$.
For $y = 0.605x+0.004$:
Substitute $x = 6.1$ into the equation: $y=0.605\times6.1 + 0.004=3.6905+0.004 = 3.6945
eq12.0$.
For $y = 0.959x + 0.041$:
Substitute $x = 6.1$ into the equation: $y=0.959\times6.1+0.041 = 5.8499+0.041=5.8909
eq12.0$.
For $y = 1.518x+0.995$:
Substitute $x = 6.1$ into the equation: $y=1.518\times6.1 + 0.995=9.2598+0.995 = 10.2548\approx10.2$ (not a great fit for this point).
For $y = 1.967x+0.984$:
Substitute $x = 6.1$ into the equation: $y=1.967\times6.1+0.984=12.0087+0.984 = 12.9927\approx12.0$ (a relatively good fit).
Let's check another point, say $(9.1,15.2)$.
For $y = 1.967x+0.984$:
Substitute $x = 9.1$ into the equation: $y=1.967\times9.1+0.984=17.9997+0.984 = 18.9837
eq15.2$.
Let's use a more formal approach. We can use the least - squares regression formula $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ 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 $\bar{y}=\frac{1}{n}\sum_{i = 1}^{n}y_i$.
However, by trial - and - error of substituting data points:
Let's use the point $(5.0,8.1)$
For $y = 0.605x+0.004$, $y=0.605\times5.0+0.004 = 3.025+0.004=3.029
eq8.1$.
For $y = 0.959x + 0.041$, $y=0.959\times5.0+0.041=4.795 + 0.041=4.836
eq8.1$.
For $y = 1.518x+0.995$, $y=1.518\times5.0+0.995=7.59+0.995 = 8.585\approx8.1$.

Answer:

$y = 1.518x + 0.995$