P1=(0,0)=(x1,y1)→x1=0, y1=0
P2=(3,-2)=(x2,y2)→x2=3, y2=-2
Slope: m=(y2-y1)/(x2-x1)
m=(-2-0)/(3-0)
m=(-2)/3
m=-2/3
Point-slope equation:
y-y1=m(x-x1)
y-0=(-2/3)(x-0)
y=-(2/3)x
Answer: The equation of the line is: y=-(2/3)x
We have:
The generic equation of the line is: y-yo = m (x-xo)
The slope is:
m = (y2-y1) / (x2-x1)
m = (- 2-0) / (3-0)
m = -2 / 3
We choose an ordered pair
(xo, yo) = (0, 0)
Substituting values:
y-0 = (- 2/3) (x-0)
Rewriting:
y = (- 2/3) x
Answer:
The equation of the line is:
y = (- 2/3) x
Answer:

Step-by-step explanation:
<u>Simple Linear Regression
</u>
It a function that represents the relationship between two or more variables in a given data set. It uses the method of the least-squares regression line which minimizes the error between the estimate function and the real data.
Let's compute the best-fit line for the data


First, we find the sums


Then, we compute the averages values


We will also compute the sums of the cross-products and the sum of the squares



We will compute Sxy and Sxx





The slope of the linear regression function is given by

The y-intercept ot the linear function is


Thus the best-fit line is

The correct option is the last one
Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027