A because one input can have only one output
i did the awnser is 173,668.18 there you goStep-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
Sin (A) = BC/AB = 21/75= 7/25
1.
5514 divided by 82 you have to find out how many times 82 goes into 551 (because it won't go into 5 or 55 now 82 goes into 551 6 times and you multiply 82 and 6 to get 492 (which is the closest you can get to 551 without going over) now you subtract 551-492 to get 59 now you carry down your 4 and add it to the end of the equation and your number changes into 594 now 82 times 7 is 574 when you subtract you get 20 now you add a decimal point and you carry down the 0 to make your number 200 now you do 82 times 2 to get 164 when you subtract you get 36 now you carry down the 0 and get 360 so you multiply 82 by 4 to get 328 so your answer is 67.24.Hope this helps. I can add a picture if you need one.
