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pickupchik [31]
3 years ago
7

Determine the level of measurement of the variable. an officer's rank in the military Group of answer choices

Mathematics
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

Ordinal

Step-by-step explanation:

Level of measurement used in statistics summarizes what statistical analysis that is possible. There exist three types of level of measurement. The nominal, ordinal and Interval/Ratio level of measurement. Here, our primary focus will be the Ordinal level of measurement.

Ordinal level of measurement indicates the position in a sequence. In the military sector, the officer's rank is said to be Ordinal. This implies that the ordinal level of measurement categorizes variables according to hierarchy or ranks with a meaningful order. Still, the intervals and differences between the variables may not be equal.

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Answer:

6 quarters and 15 dimes

Step-by-step explanation:

6 quarters = $1.50

15 dimes = $1.50

6 quarters + 15 dimes = $3.00

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Help me I need the answer ASAP ! Aundrea painted the background of a square mural on the wall using 138ft^2 of paint .which meas
Klio2033 [76]

Hey there! I'm happy to help!

Since this mural is a square, all the sides are the same. This means that you square on of the sides to get the area (multiply it by itself). If we have the area and want to find a side, we just go backwards and use something called the square root!

If we plug this into our calculators...

√138≈11.74734

We see that is closest to B. 12 ft.

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Have a wonderful day! :D

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Hi please help me I'm in the middle of a test and I don't want to get grounded
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Answer:

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Step-by-step explanation:

4 0
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Read 2 more answers
Solving a Two-Step Matrix Equation<br> Solve the equation:
Cloud [144]

Answer:

\boxed {x_{1} = 3}

\boxed {x_{2} = -4}

Step-by-step explanation:

Solve the following equation:

\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

3x_{1} + 2x_{2} + 1 = 2

-Second equation:

5x_{1} + 5x_{2} + 2 = -3

-Choose one of the two following equations, which I choose the first one, then you solve for x_{1} by isolating

3x_{1} + 2x_{2} + 1 = 2

-Subtract 1 to both sides:

3x_{1} + 2x_{2} + 1 - 1 = 2 - 1

3x_{1} + 2x_{2} = 1

-Subtract 2x_{2} to both sides:

3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1

3x_{1} = -2x_{2} + 1

-Divide both sides by 3:

3x_{1} = -2x_{2} + 1

x_{1} = \frac{1}{3} (-2x_{2} + 1)

-Multiply -2x_{2} + 1 by \frac{1}{3}:

x_{1} = \frac{1}{3} (-2x_{2} + 1)

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

-Substitute -\frac{2x_{2} + 1}{3} for x_{1} in the second equation, which is 5x_{1} + 5x_{2} + 2 = -3:

5x_{1} + 5x_{2} + 2 = -3

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

Multiply -\frac{2x_{2} + 1}{3} by 5:

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

-Combine like terms:

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

\frac{5}{3}x_{2} + \frac{11}{3} = -3

-Subtract \frac{11}{3} to both sides:

\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}

\frac{5}{3}x_{2} = -\frac{20}{3}

-Multiply both sides by \frac{5}{3}:

\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}

\boxed {x_{2} = -4}

-After you have the value of x_2, substitute for x_{2} onto this equation, which is x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}:

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

-Multiply -\frac{2}{3} and -4:

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

x_{1} = \frac{8 + 1}{3}

-Since both \frac{1}{3} and \frac{8}{3} have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

x_{1} = \frac{8 + 1}{3}

x_{1} = \frac{9}{3}

\boxed {x_{1} = 3}

5 0
3 years ago
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