<span>Suppose x=10 and y=10. what is x after evaluating the expression (y>= 10)||(x++ >10)? java
Answer is 10</span>
-5x = x + 6(1-x)
-5x = x + 6(1) + 6(-x)
-5x = x + 6 -6x
-5x = x - 6x + 6
-5x = -5x + 6
-5x + 5x = 6
0 = 6 Not equal. No solution.
(2x-5)² = (2x-1)(2x+1) -10(2x-1)
(2x-5)(2x-5) = (2x-1)(2x+1) - 20x + 10
2x(2x-5)-5(2x-5) = 2x(2x+1)-1(2x+1) - 20x + 10
4x² - 10x -10x + 25 = 4x² + 2x - 2x -1 - 20x + 10
4x² - 20x + 25 = 4x² - 20x - 1 + 10
4x² - 20x + 25 = 4x² - 20x - 9
Not equal. No solution.
Answer:
Angle ABD
Step-by-step explanation:
I think it is so because the angles are aligned already so you what has the same value.
Answer:

or

Considering ![\theta \in (0, 2\pi]](https://tex.z-dn.net/?f=%5Ctheta%20%5Cin%20%280%2C%202%5Cpi%5D)

or

Step-by-step explanation:

We have:

Therefore,

or

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or

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)