Answer:
B. (2x+4y)(4x2-8xy+16y2)
Step-by-step explanation:
hope this helps!
9514 1404 393
Answer:
in order by function: 0.6, 8, 7/3, 2, 1/2
Step-by-step explanation:
The functions are given in the form ...
f(x) = a·b^x
The "b" value is the value immediately to the left of the exponent. (It's not rocket science; it's pattern matching.) Note that the minus sign in g(x) is part of 'a', not part of 'b'.
From the top-down, the functions listed on the left have the b-values shown above.
Y=Mx +b
B is 0 because that’s where the graph intercepts the y axis.
M is found by picking any two points and finding the rise and run because m is equal rise/run
Points of choice: (-2,4) and (2,-4)
=2−1/2−1 = -8/4
m is equal -2
Y= -2X
Look at the picture below to differentiate between positive and negative slopes.
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)
Answer:
11/15
Step-by-step explanation:
Add 3 to the numerator which is 8 and that would equal 11 so your complete fraction should be 11/15.
