Supposse that the distance from the point 
to the point 
is equal to the distance from to the point 
. Then, by the formula of the distnace we must have
cancel the square root and the 
's, and then expand the parenthesis to obtain
then, simplifying we obtain
therfore we must have
this means that the points satisfying the propertie must have first component equal to 5. So we can give a lot of examples of such points: 
. The set of this points give us a straight line and the points (3,0) and (7,0) are symmetric with respect to this line.
Answer:
false
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2) P(G) = 5/8
3) P(G/E) = P(GE)/P(E)
GE = {G2, G4}
Hence P(G/E) = 2/5
4) GE = {G2, G4}
P(GE) = 2/8 = 1/4
5) P(G or E) = P(G)+P(E)-P(GE)
= 5/8 + 3/8-2/8 = 3/5
6) No there is common element as G2 and G4
Cannot be mutually exclusive
