So you should label the x axis and y axis like I did then you use the formula!
Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3
I think it’s c I had this question lol
The sum of the inner angles of any triangle is always 180°, i.e. you have
In the particular case of an equilater triangle, all three angles are the same, so
and the expression becomes
which implies 
So, if you rotate the triangle with respect to its center by 60 degrees, the triangle will map into itself. In particular, if you want point A to be mapped into point B, you have to perform a counter clockwise rotation of 60 degrees with respect to the center of the triangle.
Of course, this is equivalent to a clockwise rotation of 120 degrees.
Finally, both solutions admit periodicity: a rotation of 60+k360 degrees has the same effect of a rotation of 60 degrees, and the same goes for the 120 one (actually, this is obvisly true for any rotation!)
Answer:
Female: 11
Male: 16
Total:
(1) 34
(2) 25
(3) 100
Step-by-step explanation:
Female:
45 - 18 - 16 = 11
Male:
55 - 25 - 14 = 16
Total:
(1) 18 + 16
(2) 14 + 11 = 25
(3) 45 + 55 =
