Sorry I was wrong, I didn’t mean to get it wrong.
The common difference if there is one is the constant difference that occurs between any term and the term before it.... in this case:
There is no common difference,
dx=18,20,16,18 the difference or velocity is not constant...
d2x=2, -4,2 the acceleration is not constant...
d3x=-6,6 the thrust is not constant
Now we might be tempted to say that:
d4x=12 and say that that is constant and we COULD make a quartic equation fit all the data points, but without further data points in the sequence there is no mathematical proof that the quartic equation would produce accurate data points outside of the range given...
And solving a system of five equations for five unknowns is tedious for such a problem...a^4+bx^3+cx^2+dx+e=y
Given:
To find:
The length of QR.
Solution:
We have,
Then, 
(CPCTC)
Divide both sides by 4.
Similarity,
(CPCTC)
Divide both sides by 2.
Now,
Therefore, the correct option is d.
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)
Please show all of the expressions.
