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Nimfa-mama [501]
3 years ago
5

You take the four Aces, four $2$'s, and four $3$'s from a standard deck of 52 cards, forming a set of $12$ cards. You then deal

all $12$ cards at random to four players, so that each player gets three cards. What is the probability that each player gets an Ace, a $2$, and a $3$?

Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Answer:

Probability each player gets an ace, a $2 and a $3 = 0.0374

Step-by-step explanation:

The total number of ways to divide the card in triples among four players = 369600 ways

The total number of ways to share the cards such that no card is repeated in each triple = 13824 ways

Probability each player gets an ace, a $2 and a $3 = 13824/369600

Probability each player gets an ace a $2 and a $3 = 0.0374

Note: Further explanation is provided in the attachment.

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The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is th
Dafna11 [192]

Answer:

a + b \geq 30

b - a \geq 10

Step-by-step explanation:

Given

The sum of the two positive integer a and b is at least 30, this means the sum of the two positive integer is 30 or greater than 30, so we write the inequalities as below.

a + b \geq 30

The difference of the two integers is at least 10, if b is the greater integer then we subtract integer a from integer b, so we write the inequality as below.

b - a \geq 10

Therefore, the following system of inequalities could represent the values of two positive integers a and b.

a + b \geq 30

b - a \geq 10

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4 years ago
CAN SOMEbody please help me
tensa zangetsu [6.8K]
Https://www.calculatorsoup.com/calculators/geometry-solids/surfacearea.php
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3 years ago
Janet is drawing the path formed by the parametric equations x=2+3 sin t and y=1-1/2cos t. Which curve did she draw??
irakobra [83]
Sin²t +cos²t =1

<span> x=2+3 sin t
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</span><span>y=1-1/2cos t
y-1= - (cos t)/2 
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What is the values of the soul out ion to the inequality<br> 9 &lt; 3x + 4?
HACTEHA [7]

Answer:

x>5/3

Step-by-step explanation:

9<3x+4

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5 0
3 years ago
Howard chose a candy from a bowl with 5 chocolate candies, 4 gummy candies and 6 hard candies. What is Howard's dependent probab
Hatshy [7]

Answer:

8.9%

Step-by-step explanation:

Here, we are to calculate the probability of Howard choosing a chocolate candy followed by a gummy candy.

The probability of selecting a chocolate candy = number if chocolate candy/ total number of candy

Total number of candy = 5 + 4 + 6 = 15

Number of chocolate candy = 5

The probability of selecting a chocolate candy = 5/15 = 1/3

The probability of selecting a gummy candy = number of gummy candies/total number of candies

Number of gummy candy = 4

The probability of selecting a gummy candy = 4/15

The probability of selecting a chocolate candy before a gummy candy = 1/3 * 4/15 = 4/45 = 0.088888888889

Which is same as 8.89 percent which is 8.9% to the nearest tenth of a percent

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