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In math terms this means x divided by 3=3 so x so x means 9 so the answer is nine
we know that
For a polynomial, if
x=a is a zero of the function, then
(x−a) is a factor of the function. The term multiplicity, refers to the number of times that its associated factor appears in the polynomial.
So
In this problem
If the cubic polynomial function has zeroes at 2, 3, and 5
then
the factors are
Part a) Can any of the roots have multiplicity?
The answer is No
If a cubic polynomial function has three different zeroes
then
the multiplicity of each factor is one
For instance, the cubic polynomial function has the zeroes
each occurring once.
Part b) How can you find a function that has these roots?
To find the cubic polynomial function multiply the factors and equate to zero
so
therefore
the answer Part b) is
the cubic polynomial function is equal to
Answer:
radius: DF (it is a line segment drawn from the center of the circle to any point on the circle)
diameter: AC ( it is a straight line passing through the center of the circle and intersecting the circle at two distinct points)
chord : AB ( it is a line segment joining any two points on the circle)
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)
