darius is studying the relationship between mathematics and art. he asks friends to each draw a \typical\…

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?\n\n$y = 0.605x + 0.004$\n$y = 0.959x + 0.041$\n$y = 1.518x + 0.995$\n$y = 1.967x + 0.984$

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?\n\n$y = 0.605x + 0.004$\n$y = 0.959x + 0.041$\n$y = 1.518x + 0.995$\n$y = 1.967x + 0.984$

Answer

Explanation:

Step1: Recall the linear - regression concept

We want to find the line of best - fit for the data points ((x,y)) in the form (y = mx + b). One way is to use a graphing calculator or software to perform linear regression. Another way is to test each equation by substituting the (x) - values of the data points and seeing how close the calculated (y) - values are to the actual (y) - values. Let's take a sample point, say ((6.1,12.0)).

Step2: Test the first equation (y = 0.605x+0.004)

Substitute (x = 6.1) into (y = 0.605x + 0.004): (y=0.605\times6.1 + 0.004=3.6905+0.004 = 3.6945), which is far from (12.0).

Step3: Test the second equation (y = 0.959x+0.041)

Substitute (x = 6.1) into (y = 0.959x + 0.041): (y=0.959\times6.1+0.041 = 5.8499+0.041=5.8909), which is far from (12.0).

Step4: Test the third equation (y = 1.518x+0.995)

Substitute (x = 6.1) into (y = 1.518x + 0.995): (y=1.518\times6.1+0.995=9.2598 + 0.995=10.2548), which is relatively close.

Step5: Test the fourth equation (y = 1.967x+0.984)

Substitute (x = 6.1) into (y = 1.967x + 0.984): (y=1.967\times6.1+0.984=12.0987+0.984 = 13.0827), which is a bit far from (12.0). We can also calculate the sum of the squared errors for each equation for all data points, but by simple substitution of one point ((6.1,12.0)), the equation (y = 1.518x + 0.995) gives the closest result among the four.

Answer:

(y = 1.518x + 0.995)