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AleksandrR [38]
1 year ago
12

The method of least squares of residuals is commonly used to get the best fit with linear regression. The reason why the absolut

e value of residual's (|y- ypred|) is not used is that?
Mathematics
1 answer:
Fudgin [204]1 year ago
7 0

The method of least squares of residuals is commonly used to get the best fit with linear regression. The reason why the absolute value of residual's (|y- ypred|) is not used is as follows :

Although least squares purposefully square the error in the belief that it will respond better when values deviate from your projected value, you could run something similar if you so desired.

Consider the following scenario: You plot your data's real values on a graph, and you observe a straight line of numbers with a few outliers that have greater values than you would expect. You could be tempted to believe that the best fit is a straight line along the values. Because the real pattern must take into account all the values that occurred, least squares adapt more to the outliers than other values and purposefully squares values, sort of assuming that the straight line of actual values was a coincidence.

To learn more about least squares here:

brainly.com/question/2141008

#SPJ4

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Answer:

A circle has a circumference of 7,805 units.

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sladkih [1.3K]

D is halfway between A and B

so the coordinates of D are (2,2)

E is halfway between A and C so the coordinates of E are (-1,1)

now you need to find the gradient/slope of DE and BC using the formula:

\frac{y2 - y1}{x2 - x1}

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>D</u><u>E</u><u>:</u><u> </u></h3>

SUB IN COORDINATES OF D AND E

\frac{1 - 2}{ - 1 - 2}

therefore the gradient of DE is 1/3.

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>

<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>

<em>\frac{ - 2 - 0}{ - 3 - 3}</em>

therefore the gradient of BC is -2/-6 which simplifies to 1/3.

<h3>therefore, BC and DE are parallel as they both have a gradient/slope of 1/3 and parallel lines have the same gradient</h3>

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Its octagon
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a

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