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 = ![^{52}C_9](https://tex.z-dn.net/?f=%5E%7B52%7DC_9)
Now , the combination of 9 cards such that exactly 6 of them are face cards = ![^{12}C_{6}\times^{40}C_3](https://tex.z-dn.net/?f=%5E%7B12%7DC_%7B6%7D%5Ctimes%5E%7B40%7DC_3)
Now , the probability that the hand contains exactly 6 face cards will be :-
![\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5E%7B12%7DC_%7B6%7D%5Ctimes%5E%7B40%7DC_3%7D%7B%5E%7B52%7DC_9%7D)
![=\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: d
Step-by-step explanation:because if look at the problem 8-1 is 7 and 3/4-2/3 is around 1 so it would be 6
Solution: We are given:
![Mean =78, Standard-deviation =6.3](https://tex.z-dn.net/?f=Mean%20%3D78%2C%20Standard-deviation%20%3D6.3)
We know that a usual values of the test scores falls within 2 standard deviation from the mean.
Therefore, the minimum usual test score is:
![Mean - 2 Standard-deviation](https://tex.z-dn.net/?f=Mean%20-%202%20Standard-deviation)
![78-2 \times 6.3](https://tex.z-dn.net/?f=78-2%20%5Ctimes%206.3)
![78-12.6](https://tex.z-dn.net/?f=78-12.6)
![65.4](https://tex.z-dn.net/?f=65.4)
The maximum usual test score is:
![Mean + 2 Standard-deviation](https://tex.z-dn.net/?f=Mean%20%2B%202%20Standard-deviation)
![78+2 \times 6.3](https://tex.z-dn.net/?f=78%2B2%20%5Ctimes%206.3)
![78+12.6](https://tex.z-dn.net/?f=78%2B12.6)
![90.6](https://tex.z-dn.net/?f=90.6)
Therefore, the minimum and maximum “usual” values of the test scores are:
65.4 and 90.6
7!/3!
7! = 7 X 6 X 5 X 4 X 3 ...
3! = 3 X 2 x 1
since the 3 X 2 X 1 part are repeated in both the numerator and denominator, we can divide them out
7!/3! = 7 X 6 X 5 X 4 = 42 X 20 = b. 840
15h + 80 = 140
If you wanted to solve how many hours he worked:
15h + 80 = 140
15h = 60
h = 4.