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Rainbow [258]
3 years ago
9

POZ HELP FAST I NEER HELP

Mathematics
2 answers:
kupik [55]3 years ago
7 0

Answer: \frac{14}{1}

Step-by-step explanation:

Simplify \frac{70}{5} = \frac{14}{1}

Rama09 [41]3 years ago
7 0

Answer:

14/1

Step-by-step explanation:

70 divided by 5

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kiruha [24]
Is it just x=b-cd
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cd=b-x
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6 0
3 years ago
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Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI
cricket20 [7]

Answer:

The system has infinitely many solutions

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

  1. Switch any two rows
  2. Multiply a row by a nonzero constant
  3. Add one row to another

To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

7 0
3 years ago
The median number of magazine appearances made by 7 models is 5. The range of number of magazine appearances by those models is
sattari [20]

Answer:not enough information

Step-by-step explanation:

7 0
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Movies Tonight is a typical video and DVD movie rental outlet for home-viewing customers. During the weeknight evenings, custome
stepladder [879]

Answer:

the probability having no customers are in the system is 0.375

Step-by-step explanation:

The computation of the probability having no customers are in the system is as follows;

Given that the arrival rate is 1.25 customers per minute

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Based on the above information, the probability is

=P_0\\\\=(1-\frac{\lambda }{\mu })\\\\=(1-\frac{1.25}{2})\\\\=1-0.625\\\\= 0.375

Hence, the probability having no customers are in the system is 0.375

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24 inches per minute converts to how many miles per day
12345 [234]

Answer:

0.545455

Step-by-step explanation:

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