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 equation of a line

The equation of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. We can use the method of least - squares regression (or a quick estimate by choosing two points and calculating slope approximately). Let's use a point - by - point check.

Step2: Take the first point $(6.1,12.0)$

For option 1: $y=0.605x + 0.004$, when $x = 6.1$, $y=0.605\times6.1+0.004=3.6905 + 0.004=3.6945
eq12.0$.
For option 2: $y = 0.959x+0.041$, when $x = 6.1$, $y=0.959\times6.1 + 0.041=5.8499+0.041=5.8909
eq12.0$.
For option 3: $y=1.518x + 0.995$, when $x = 6.1$, $y=1.518\times6.1+0.995=9.2598 + 0.995=10.2548\approx10.3$. When $x = 5.0$, $y=1.518\times5.0+0.995=7.59+0.995 = 8.585\approx8.6$.
For option 4: $y=1.967x+0.984$, when $x = 6.1$, $y=1.967\times6.1+0.984=12.0987+0.984=13.0827
eq12.0$.
We can also calculate the slope between two points. For example, between $(5.0,8.1)$ and $(6.1,12.0)$ the slope $m=\frac{12.0 - 8.1}{6.1 - 5.0}=\frac{3.9}{1.1}\approx3.545$. But a more accurate way is to use linear regression. However, by substituting points into the equations of the lines, we find that the line $y = 1.518x+0.995$ gives values that are closest to the given data points.

Answer:

$y = 1.518x+0.995$