Answer:
(x + 6)(x + 13)
Step-by-step explanation:
Given
x² + 19x + 78
Consider the factors of the constant term (+ 78) whuch sum to give the coefficient of the x- term (+ 19)
The factors are + 6 and + 13 , since
6 ×13 = + 78 and 6 + 13 = + 19 , then
x² + 19x + 78 = (x + 6)(x + 13) ← in factored form
Answer:
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Answer:
- 1/2
Step-by-step explanation:
We use the slope-intercept form to figure out our slope.
y = mx + b
Get y alone:
x + 2y = -8
2y = -x - 8
y = -x/2 - 4
Since m is the slope in y = mx + b, we can see that -1/2 is in the place of m, therefore our slope is -1/2
82,000 bc 8% of 50,000 is 4000 and 4000 times 8 is 32,000 + 50000
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)
