There would be 2 raised to c elements in the power set. Elements of the
power set are formed by making a binary decision on each element of the input
set: include or do not include. Since there are two choices for each element of
the input set, and there are c elements in the input set, there are 2^c possible
arrangements of these input elements.
Answer:
1200
Step-by-step explanation:
We’ll need a number that is evenly dividable by all three values: 8,12 & 15. We can develop the generic common multiple by multiplying all three numbers and get 1440. But is that the smallest common multiple? The quick and dirty way to factor the number is to divide it by it’s various elements and see what works. In this case 1440/12=120 which is the smallest common multiple of the three numbers. The smallest length is therefore 120.
In order to find how many bricks just divide (120x120x120)/(8x12x15)=1200 bricks
Answer:
Emily , don't put your whole name in the questions :DDDD, it's at the top of the page, anyway, Eloper .. :DDD
Step-by-step explanation:
It's arithmetic b/c you're just adding 1
arithmetic
1
1
are the answers , the last one is b/c that's the common difference.
Answer: 
or 
.
Step-by-step explanation:
Given : A poker hand consisting of 9 cards is dealt from a standard deck of 52 cards.
The total number of cards in a deck 52
Number of faces cards in a deck = 12
Number of cards not face cards = 40
The total number of combinations of drawing 9 cards out of 52 cards = 
Now , the combination of 9 cards such that exactly 6 of them are face cards = 
Now , the probability that the hand contains exactly 6 face cards will be :-
![=\dfrac{\dfrac{12!}{6!6!}\times\dfrac{40!}{3!37!}}{\dfrac{52!}{9!\times43!}}\ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{228}{91885}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdfrac%7B12%21%7D%7B6%216%21%7D%5Ctimes%5Cdfrac%7B40%21%7D%7B3%2137%21%7D%7D%7B%5Cdfrac%7B52%21%7D%7B9%21%5Ctimes43%21%7D%7D%5C%20%5C%20%5B%5Cbecause%5C%20%5EnC_r%3D%5Cdfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D%5D%5C%5C%5C%5C%3D%5Cdfrac%7B228%7D%7B91885%7D)
Hence, the probability that the hand contains exactly 6 face cards. is .
Answer:
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