It shouldn't be too tough to find one of those, seeing that there are
an infinite number of them.
To create one, take any integer, positive or negative, and multiply it by itself.
Here are a few to put you in the mood:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, ...
784, 841, 900, 1024, 1225, 1600, 2500, 3600, 4900, 10000, 1 million, ...
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.
1
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)
What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note:
A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with
13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant,
and you keep each card as it is dealt -- it's not returned to the deck.)
The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken
from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and
there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the
first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for
the other three. The probability of the royal flush is therefore the product of these numbers, or
5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
Answer:
91
Step-by-step explanation:-5, 1, 7, 13
the common diference is : d = ( 13)-(7)=(7)-(1)=(1)-(-5)=6
the firt term is A1 = -5
the n-ieme term is : An = A1 +(n-1)d
An = -5+6(n-1)
An = 6n-11
the 17th term in the arithmetic sequence is : A17 = 6(17)-11 =91
Read more on Brainly.com - brainly.com/question/15374982#readmore
Answer:
21 ounces of flour
Step-by-step explanation:
10/3 = 70/x
isolate the so that it is on the top, so multiply by x on the right and on the left
this gives you:
x*10/3=70
try to isolate x and multiply by 3 to get rid of the denominator
x*10=70*3
x*10= 210
finally isolate x by dividing by 10 and getting rid of the 10 multiplying with it
x= 210/10
x=21
you need 21 ounces of flour if you have 70 ounces of sugar you can double check by making sure the ratio of 10/3 is equal to 70/21 and of course they both equal 3.33 repeating, therefore your answer is correct Hope that helps
