Answer: The answer is D. .625
Step-by-step explanation: It terminates, meaning that it doesn't repeat itself
Could you please provide the equation and the rest of the question?
Answer:
The length of the line segment AC is equal to 14
Step-by-step explanation:
The triangle above is an isosceles triangle, In an Isosceles triangle the two angles; B and C are the same, hence the two sides; AB and AC are also the same.
AB=2x and AC= 3x - 7
AB = AC
which implies;
2x = 3x - 7
subtract 3x from both-side of the equation
2x - 3x = 3x -3x -7
-x = -7
Multiply through by -1
x = 7
But we were ask to find the the length of the line segment AC
AC = 3x - 7
substituting x = 7 into the above equation will yield;
AC = 3(7) - 7 = 21 - 7 =14
Therefore the length of the line segment AC is equal to 14
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)
