Which ordered pairs could be points on a line parallel to the line that contains (3, 4) and (–2, 2)? Check all that apply. (–2, –5) and (–7, –3) (–1, 1) and (–6, –1) (0, 0) and (2, 5) (1, 0) and (6, 2) (3, 0) and (8, 2)

(–1, 1) and (–6, –1)

(1, 0) and (6, 2)

(3, 0) and (8, 2)

Step-by-step explanation:

Slope of line having points (x1, y1) and (x2,y2) is given by (y2-y1)/(x2-x1)

Two lines with slope m1 and m2 are parallel if there slopes are equal i.e m1 = m2.

Slope of line  that contains (3, 4) and (–2, 2)

m = (2-4)/(-2-3) = -2/-5 = 2/5

To find if other pair of lines lie on line parallel to  line  that contains (3, 4) and (–2, 2), we need to find slope for other pair of points , if those pairs have slopes as 2/5 then its clear that they lie on line parallel to  line  that contains (3, 4) and (–2, 2) else not.

Lets find slope of all the pairs

(–2, –5) and (–7, –3)

m1 = -3 -(-5)/-7-(-2) = -3+5/-7+2

m1 = 2/-5 = -2/5

as  -2/5 is not equal to 2/5, the mentioned points do not  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(–1, 1) and (–6, –1)

m2 = -1 -1)/-6-(-1) = -2/-6+1

m2 = -2/-5 = 2/5

as  2/5 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(0, 0) and (2, 5)

m3 = 5 -0)/2-0

m3 = 5/2

as   5/2 is not equal to 2/5, the mentioned points do not lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(1, 0) and (6, 2)

m4 = 2 -0)/6-1

m4 = 2/5

as  m4

As m4 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(3, 0) and (8, 2)

m5 = 2 -0)/8-3

m5 = 2/5

as  m5 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

Thus, pair of points which lie on  lie on line parallel to  line  that contains (3, 4) and (–2, 2) are

(–1, 1) and (–6, –1)

(1, 0) and (6, 2)

(3, 0) and (8, 2)

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560