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Gekata [30.6K]
3 years ago
13

Common denominator of 81 and 63

Mathematics
1 answer:
fomenos3 years ago
7 0

9

9 x 9 = 81

9 x 7 = 63

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Use matrices and elementary row to solve the following system:
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I assume the first equation is supposed to be

5x-3y+2z=13

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5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

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\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

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\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

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\boxed{x=1,y=-2,z=1}

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3 years ago
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.

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