Solution:
The probability of an event is expressed as
In a pack of 52 cards, we have
Thus, we have the probability to be evaluated as
Answer:
C. y = 7x + 29 .
Step-by-step explanation:
First find the slope of the given line:
x - 7y = 12
7y = x - 12
y = 1/7 x - 12/7
The slope = 1/7 so the slope of the line perpendicular to this line is
- 1 / -1/7 = 7.
Finding the equation of the required line:
y - y1 = m(x - x1)
y1 = 8 , m = 7 and x1 = -3:
y - 8 = 7(x + 3)
y = 7x + 21 + 8
y = 7x + 29 answer.
Answer:
The statistical conclusion is:
a) Your results are significant; one's willingness to wear masks is not independent of one's age.
Step-by-step explanation:
From this experiment,
The null hypothesis is established as:
H 0 = one's willingness to wear masks is independent of one's age.
The alternate hypothesis is established as:
H 1 = one's willingness to wear masks is not independent of one's age.
In statistics, if the absolute value of the test statistic is greater than the critical value, we declare statistical significance and reject the null hypothesis.
Since the test statistic = 16.5, which is higher than the critical value of 7.82, this establishes statistical significance. Thus, the null hypothesis is rejected.
Therefore, the statistical conclusion is that one's willingness to wear masks is not independent of one's age.
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)
5 2/3-2 1/4=miles ran on Monday
5 8/12-2 3/12=miles ran on Monday
3 5/12=miles ran on Monday
