Answer:
Line BD can't be represent by the image.
Explanation:
Point: Point is a dot or vertex of any figure or diagram. It represent by capital alphabet. So, B would be a point.
Line: A line is a straight path with both endless. It means we can extends line from both end. But in case of BD we can't extend from point B. So, it won't be line.
Ray: A ray is straight path with one endless. It means we can't extend line from one end and extend from another end. But we can extend AC from both end. It won't be a ray.
Line Segment: A line segement is straight path with end from both end. We can't extend from either end of the line. We can say CD would be line segment.
Option 2 is correct.
Apply Pythagoras:
length = sqrt( (10--4)² + (6-3)² ) = sqrt(205)
You didn’t add a screenshot. I’m sorry for answering but someone else will if I don’t.
Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way
Step-by-step explanation:
- From a standard deck of cards, one card is drawn. What is the probability that the card is black and a
jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
- A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen
or an ace.
P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13
- WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces?
P(AA) = (4/52)(3/51) = 1/221.
- WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king?
P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed.
- WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the
probability of drawing the first queen which is 4/52.
- The probability of drawing the second queen is also 4/52 and the third is 4/52.
- We multiply these three individual probabilities together to get P(QQQ) =
- P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
- Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
36
Step-by-step explanation:
360/2=180
180/5=36
