Slope = (2+5)/(-3-1) = 7/-4 = -7/4
answer is B -7/4
To answer this question, you do 5 2/3 divided by 1/2. First, change 5 2/3 to 17/3. Then, use the reciprocal method by doing 17/3 times 2/1, which gives you 34/3. That’s 11.3 repeating, so they can make 11 whole snowmen.
Locate 1 on the x axis. This is the horizontal number line.
Draw a vertical line through 1 on the x axis. Extend this vertical line as far up and down as you can.
Notice how the vertical line crosses the blue curve. Mark this point. Then draw a horizontal line from this point to the y axis. The horizontal line will touch -4 on the y axis. So that means the point (1,-4) is on the curve.
If the input it is x = 1, then the output is y = -4
So that's why the answer is choice A
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)
