Answer:
f(x) = 4 when x is 8
Step-by-step explanation:
Domain is the set of x values that make the function defined. Allowed x values for the function (mapping).
The Range is the set of y values that make the function defined. Allowed y values for the function (mapping).
- Whenever we need to find f(a), suppose, then we look for "a" in the domain and see its corresponding value mapping in the range.
- Whenever we will be given a value for f(x) = a, suppose, and we have to find "x", we look at the value a in the range and find corresponding x value in the domain.
Firstly, we need f(4), so we look for "4" in domain and see which number it corresponds to in range.
That is 
Thus,
Next,
We want "x" value that gives us a "y" value of 4. We look for "4" in the range and see which value it corresponds to. That is "8". So,
f(8) = 4
What you want is P(6∩1) or P(1∩6) or P(2∩5) or P(5∩2) or P(3∩4) or P(4∩3).
The events of rolling the dice are independent (i.e. they don't affect one another) so:
E.g.
P(6∩1) = P(6) * P(1)
P(2∩5) = P(2) * P(5)
The probability of getting a given number on a roll is 1/6 for both dice.
So:
P(6∩1) = 1/6 * 1/6 = 1/36
This is the same for any arrangement of numbers you could get from rolling two dice.
So, we can see that there are 6 arrangements of numbers that will give a sum of 7 and so that is 6 * 1/36 = 6/36 = 1/6
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)
183.9 millimeters is the answer
Pay attention in math class! in higher grade levels, this is gonna show up EVERYDAY! no joke
In these types of problems, you need to use PEMDAS also.
((4)+3)+(13(-2)-5)
(7)+(-31)
7-31
-24
<u><em>C</em></u>
