The factors of 7are -1 and 7 or 1 and -7, the factors of 14 are 1, 2, 7, and 14, or -1, -2, -7,-14. so the list of potential zeros are: 1/1, 1/2, 1/7, 1/14, 7/1,7/2, 7/7, 7/14, which can be simplified into 1, 1/2,1/7, 1/14, 7, 7/2
add the negative ones: -1, -1/2,-1/7, -1/14, -7, -7/2
I believe there are a total of 12 potential zeros
reference:
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml
Answer:
Is it 2?
Step-by-step explanation:
Answer:
$150
Step-by-step explanation:
convert the percent to decimal, 0.40, then multiply it by 250 to get 100, then subtract, 250-100=$150
Answer is <span>16y + 7z
</span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>−<span>4z
</span></span><span>=<span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+</span>−<span>4z
</span></span></span>Combine Like Terms:
<span>=<span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+<span>−<span>4z
</span></span></span></span><span>=<span><span>(<span><span>9y</span>+<span>7y</span></span>)</span>+<span>(<span><span>11z</span>+<span>−<span>4z</span></span></span>)
</span></span></span><span>=<span><span>16y</span>+<span>7z</span></span></span><span>
</span>
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.
1
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)
What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note:
A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with
13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant,
and you keep each card as it is dealt -- it's not returned to the deck.)
The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken
from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and
there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the
first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for
the other three. The probability of the royal flush is therefore the product of these numbers, or
5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
