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nekit [7.7K]
4 years ago
9

How many times can 3 go into 6.

Mathematics
2 answers:
IRINA_888 [86]4 years ago
8 0
2, dos, two hahah 3x2 = 6
taurus [48]4 years ago
5 0
3 can go into 6 two times
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Z times 231 equals 256 write as an equation
oksano4ka [1.4K]

Answer:

231z = 256

Step-by-step explanation:

231z = 256.  That's all you have to do.  If you wish, divide both sides by 231 to find the value of z.

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What is the area of a triangle measuring 2.2 width and 9.9 length
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I will assume that by 'width' you mean the base of the triangle. Then, the area would be 10.89cm^2
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The game of clue involves 6 suspects, 6 weapons, and 9 rooms. one of each is randomly chosen and the object of the game is to gu
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Part A:

Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.

The number of ways that one of each is randomly chosen is given by:

^6C_1\times{ ^6C_1}\times{ ^9C_1}=6\times6\times9=324

Therefore, the number of solutions possible is 324.



Part B:

Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

Let x denote the number of solutions that are possible after that player observes his or her three cards, then:

x={ ^{6-s}C_1}\times{ ^{6-w}C_1}\times{ ^{9-r}C_1}=(6-s)(6-w)(9-r)

Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).



Part C:

The expected value E(x) of a data set x_i with probabilities p(x_i) is given by E(x)=\Sigma xp(x)

There are </span>^{3+3-1}C_{3-1}={ ^5C_2}=10 possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

Thus the expected value is given by

E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\  \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}

+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\  \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\  \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\  \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\  \\ =\frac{1}{1,330}(264,384) \\  \\ =\bold{198.78}
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You roll a red number cube and a blue number cube. What is the probability you roll a red 2 and a blue 2?
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Answer:

The answer will be 4

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15,281

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