What is the equation of the line of best fit that Samir drew? OA. y = -x + 12 OB. O C. y = - + 18 y = -x + 12 y = -x + 18 OD.

 Choice A

y = (-2/3)x + 12

Work Shown:

We will only focus on the two points that are on the regression line.

Those two points are (3,10) and (12,4)

Let (x1,y1) = (3,10) and (x2,y2) = (12,4)

Find the slope of the line through (x1,y1) = (3,10) and (x2,y2) = (12,4)

m = (y2 - y1)/(x2 - x1)

m = (4 - 10)/(12 - 3)

m = -6/9

m = -2/3

Plug m = -2/3 and (x1,y1) = (3,10) into the point slope formula

Solve for y.

y - y1 = m(x - x1)

y - 10 = (-2/3)(x - 3)

y - 10 = (-2/3)x + (-2/3)(-3) ... distribute

y - 10 = (-2/3)x + 2

y - 10 + 10 = (-2/3)x + 2 + 10 add 10 to both sides

y = (-2/3)x + 12

This matches up with what the graph shows because the regression line intersects 12 on the y axis, so the y intercept is 12.

y = (-2/3)x + 12 is in the form y = mx+b with m = -2/3 as the slope and b = 12 as the y intercept.

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12