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Debora [2.8K]
3 years ago
5

Maria was foing a assroom survey. She asked the girls in the class how many siblings they had and recorded the results: 1,2,3,6,

7,9,9,9,10. Determine the mean (rounded to the nearest tenth) the median, mode, and range of the results
Mathematics
2 answers:
GenaCL600 [577]3 years ago
6 0

Mean: The sum of all the numbers, divided by the amount of numbers in the set.

56 / 9 = 6.2

Median: The middle value of a set.

7

Mode: The value that occurs most frequently.

9

Range: The highest number of the set subtracted by the lowest number.

10 - 1 = 9

Hope this helps you!


umka21 [38]3 years ago
3 0
Median: 7 because it is in the very middle of the numbers

Mean: add up all the numbers and divide by how many numbers there are. So the mean should be 56÷9=6.2. 6.2(2 is repeating) is the mean

Range: the range is the difference between the highest and lowest numbers. so that means the range is 10-1=9. the range is 9
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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}

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Given the diagram below, what statement can you not make?
Vaselesa [24]

Answer:

b, 5=3

Step-by-step explanation:

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2 years ago
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Answer: B, D, and E

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What is the common ratio between successive terms in the sequence?
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It's 1/3.

Step-by-step explanation:

Common ratio = 9/27 = 1/3.

3 / 9 = 1/3

1 / 3 = 1/3   and so on.

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