Answer:
That would be (2,3)
Step-by-step explanation:
When we're moving left in the coordinate plane, we are subtracting the x-coordinate by how much distance we move.
In this case, 7-5=2, so the new x-coordinate is 2, but the y-coordinate doesn't change since we're going horizontally and not vertically.
Answer:
So the number of total combinations is 35.
Step-by-step explanation:
We know that Ellen must take 4 courses this semester. She has a list of 3 math courses and 4 science courses.
Therefore, she have total 7 courses.
So, we calculate the number of combinations to choose 4 out of 7 courses.
We get:
So the number of total combinations is 35.
<span>books to bookmarks
books : bookmarks
10 : 1
10 * 3 : 1 * 3
30 : 3
3 bookmarks
Since the book store sold 30 books they also sold 3 bookmarks.
Answer: 3 bookmarks</span>
Answer: A, C, and D all round to 5.6
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)
