Answer:
30.72
Step-by-step explanation:
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
Answer:
Yes
Step-by-step explanation:
The property used in #20 was called 'the distributive property'.
This property gets rid of the parenthesis and also gets the expression more simple.
x(y+b)
x*y + x*b ⇔ Notice the change here?
Answer would be: Distributive property
Answer:
4 1/4
Step-by-step explanation:
2 1/2 + 1 1/4 + 1/2 = 5/2 + 5/4 + 1/2
LCM = 4
5/2 x 2/2 = 10/4
5/4 = 5/4
1/2 x 2/2 = 2/4
10/4 + 5/4 + 2/4 = 17/4
= 4 1/4
Therefore he put 4 1/4 cups altogether.