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Marrrta [24]
2 years ago
7

The least-squares regression line is ________.

Mathematics
1 answer:
Murljashka [212]2 years ago
5 0

Answer:

a. the line that passes through the most data points.

Step-by-step explanation:

Regression analysis, is used to draw the line of‘ best fit’ through co-ordinates on a graph. The techniques used enable a mathematical equation of the straight line form y=mx+c to be deduced for a given set of co-ordinate values, the line being such that the sum of the deviations of the co-ordinate values from the line is a minimum, i.e.

The least-squares regression lines  is the line of best fit

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Step-by-step explanation:

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31. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
Whitepunk [10]

Answer:

The required linear equation satisfying the given conditions f(-1)=4 and f(5)=1 is $y=\frac{-1}{2} x+\frac{7}{2}$

Step-by-step explanation:

It is given that f(-1)=4 and f(5)=1.

It is required to find out a linear equation satisfying the conditions f(-1)=4

and f(5)=1. linear equation of the line in the form

$$\left(y-y_{2}\right)=m\left(x-x_{2}\right)$$

Step 1 of 4

Observe, f(-1)=4 gives the point (-1,4)

And f(5)=1 gives the point (5,1).

This means that the function f(x) satisfies the points (-1,4) and (5,1).

Step 2 of 4

Now find out the slope of a line passing through the points (-1,4) and (5,1),

$$\begin{aligned}&m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\&m=\frac{1-4}{5-(-1)} \\&m=\frac{-3}{5+1} \\&m=\frac{-3}{6} \\&m=\frac{-1}{2}\end{aligned}$$

Step 3 of 4

Now use the slope $m=\frac{-1}{2}$ and use one of the two given points and write the equation in point-slope form:

$(y-1)=\frac{-1}{2}(x-5)$

Distribute $\frac{-1}{2}$,

$y-1=\frac{-1}{2} x+\frac{5}{2}$

Step 4 of 4

This linear function can be written in the slope-intercept form by adding 1 on both sides,

$$\begin{aligned}&y-1+1=\frac{-1}{2} x+\frac{5}{2}+1 \\&y=\frac{-1}{2} x+\frac{5}{2}+\frac{2}{2} \\&y=\frac{-1}{2} x+\frac{7}{2}\end{aligned}$$

So, this is the required linear equation.

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