Answer:
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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)
SI = PRT /100
1. SI = 400*5*1/100
= 2000/100
= $ 20
2. SI = 1000*8.5*3/100
= 25.5*1000/100
= $ 255
3. SI = 200*9*1/2*100
= 900/100
= $9
4. SI = 20000*12*3/100
= 200*12*3
= 7200
5. SI = 1200*4.5*6/12*100
= 4.5*6
= $27
Amount to pay = Principal + SI
= 1200+27
= $1227 ( But this answer isn't there in the options )
Look at your picture
notice the coordinates for the center of the circle, the x,y or h,k values
notice how long the radius is
thus
Answer::2,075,673,600 batting orders may occur.
Step-by-step explanation: A baseball team has 4 pitchers, who only pitch, and 12 other players, all of whom can play any position other then pitcher.