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DerKrebs [107]
3 years ago
13

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

be the mean y-coordinate of these points.
Then ( ¯ x , ¯ y ) =

Now consider a line through the point ( ¯ x , ¯ y ) that has slope m.

Add up the squares of the vertical distances between the line and these points (your result will be a quadratic function in terms of the variable m ).

What is the value of m that makes this total as small as possible?

Mathematics
1 answer:
loris [4]3 years ago
5 0

Answer:

m=\dfrac{3}{2}

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}

<u>1) Finding (\overline{x},\overline{y})</u>

  • Average of the x coordinates:

\overline{x} = \dfrac{1+2+3}{3}

\overline{x} = 2

  • Average of the y coordinates:

similarly for y

\overline{y} = \dfrac{3+3+6}{3}

\overline{y} = 4

<u>2) Finding the line through (\overline{x},\overline{y}) with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

(y-y_1)=m(x-x_1)

in our case this will be

(y-\overline{y})=m(x-\overline{x})

(y-4)=m(x-2)

y=mx-2m+4

this is our equation of the line!

<u>3) Find the squared vertical distances between this line and the three points.</u>

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.  

  • Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

y=m(1)-2m+4

y=-m+4

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

d_1=3-(-m+4)

d_1=m-1

finally, as asked, we'll square the distance

(d_1)^2=(m-1)^2

  • Distance from point (2,3)

we'll do the same as above here:

y=m(2)-2m+4

y=4

vertical distance between the two points: (2,3) and (2,4)

d_2=3-4

d_2=-1

squaring:

(d_2)^2=1

  • Distance from point (3,6)

y=m(3)-2m+4

y=m+4

vertical distance between the two points: (3,6) and (3,m+4)

d_3=6-(m+4)

d_3=2-m

squaring:

(d_3)^2=(2-m)^2

3) Add up all the squared distances, we'll call this value R.

R=(d_1)^2+(d_2)^2+(d_3)^2

R=(m-1)^2+4+(2-m)^2

<u>4) Find the value of m that makes R minimum.</u>

Looking at the equation above, we can tell that R is a function of m:

R(m)=(m-1)^2+4+(2-m)^2

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)

\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)

now to find the minimum value we'll just use a condition that \dfrac{dR}{dm}=0

0=2(m-1)+2(2-m)(-1)

now solve for m:

0=2m-2-4+2m

m=\dfrac{3}{2}

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

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PLEASE HELPPPP ILL MARK BRAINLIEST
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7 0
2 years ago
Some people took part in a game. The frequency shows information about their scores.
Marizza181 [45]

Answer:

The mean = 17.54

Step-by-step explanation:

Form a table as below;

<u>Interval         Frequency{f}      Midpoint of frequency{x}          fx    </u>

1-7                    19                         4                                            76

8-10                  14                         9                                            126

11-15                  15                         13                                           195

16-20                20                        18                                           360

21-35                 10                        28                                          280

36-50                13                         43                                          559

Sum                   91                                                                        1596

The mean  of grouped data = Sum of { Interval Midpoint * Frequency } /Sum of frequency

The mean= 1596 / 91

The mean = 17.54

5 0
3 years ago
Find the values of x and y.
DENIUS [597]
X = 20 degrees
Y = 30
This is acute triangle so all the sides and angles are the same. Subtract 16 from 46 and you receive 30 for y. For the corresponding angle, subtract 80 from 180. Then add 60 to that and subtract that from 180 for x.
6 0
2 years ago
Which of the following is the sum of both solutions of the equation <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Bx-42%3D0"
AysviL [449]

Answer: x=6 or x=-7

Step-by-step explanation:

Step 1: Factor left side of equation: (x-6)(x+7)=0

Step 2: Set factors equal to 0: x-6=0 or x+7=0

Get your answer: x=6 or x=-7

5 0
3 years ago
4-9 ik this is probably easy but I’m stuck so ue
Lady bird [3.3K]
4-9= -5 that is the answer
4 0
3 years ago
Read 2 more answers
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