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Mama L [17]
3 years ago
7

I need this answer and FAST! This is the problem: 4=8-3h

Mathematics
1 answer:
Sergio [31]3 years ago
3 0
Rememeber, you can do anything to an equation as long as you do it to both sides


4=8-3h
minus 8 both sides
-4=-3h
divide both sides by -3
-4/-3=h
4/3=h
h=4/3 or about 1.333333333333333333333333333
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A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected
Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities =\frac{4}{33}  =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let x_1, x_2, x_3 and x_4 be four given positive integers and let x_1+x_2+x_3+x_4= N.

A random variable X is said to have hypergeometric distribution with parameter x_1, x_2, x_3 , x_4  and n.

The probability mass function

f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right)  }{\left(\begin{array}{c}N\\n\end{array}\right) }

Here a_1+a_2+a_3+a_4=n

{\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}

Given that, a foreign club is made of  2 Canadian  members, 3 Japanese  members, 5 Italian  members and 2 Germans  members.

x_1=2, x_2=3, x_3 =5 and x_4=2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

a_1=1, a_2=1,a_3=1 , a_4=1, n=4

The required probability is

=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right)  }{\left(\begin{array}{c}12\\4\end{array}\right) }

=\frac{2\times 3\times 5\times 2}{495}

=\frac{4}{33}

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)

=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right)  }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right)  }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right)  }{\left(\begin{array}{c}12\\4\end{array}\right) }

=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}

=\frac{6+12+6}{495}

=\frac{8}{165}

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

6 0
3 years ago
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