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Bogdan [553]
2 years ago
10

In Problems 1-8, determine the level of measurement of each variable.

Mathematics
1 answer:
lbvjy [14]2 years ago
3 0

The level of measurement of each given variable are:

1. Ordinal

2. Nominal

3. Ratio

4. Interval

5. Ordinal

6. Nominal

7. Ratio

8. Interval

Level of measurement is used in assigning measurement to variables depending on their attributes.

There are basically four (4) levels of measurement (see image in the attachment):

1. <u>Nominal:</u> Here, values are assigned to variables just for naming and identification sake. It is also used for categorization.

  • Examples of variables that fall under the measurement are: Favorite movie, Eye Color.

<u>2. Ordinal:</u> This level of measurement show difference between variables and the direction of the difference. In order words, it shows magnitude or rank among variables.

  • Examples of such variables that fall under this are: highest degree conferred, birth order among siblings in a family.

<u>3. Interval Scale:</u> this third level of measurement shows magnitude, a known equal difference between variables can be ascertain. However, this type of measurement has <em>no true zero</em> point.

  • Examples of the variables that fall here include: Monthly temperatures, year of birth of college students

4. Ratio Scale: This scale of measurement has a "true zero". It also has every property of the interval scale.

  • Examples are: ages of children, volume of water used.

Therefore, the level of measurement of each given variable are:

1. Ordinal

2. Nominal

3. Ratio

4. Interval

5. Ordinal

6. Nominal

7. Ratio

8. Interval

Learn more about level of measurement here:

brainly.com/question/20816026

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butalik [34]

Answer:

see below

Step-by-step explanation:

<h3>Proposition:</h3>

Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE

<h3>Proof:</h3>

1)The lines AD and BC are parallel and AC their transversal therefore,

\displaystyle  \angle DAC =  \angle ACB \\  \ \qquad [\text{ alternate angles theorem}]

2)The lines AB and DC are parallel and BD their transversal therefore,

\displaystyle  \angle BD C=  \angle ABD \\  \ \qquad [\text{ alternate angles theorem}]

3)now in triangle ∆AEB and ∆CED

  • \displaystyle \angle EAD=\angle ECB
  • \angle EDA=\angle EBC
  • \displaystyle AD=BC

therefore,

\displaystyle  \Delta AEB  \cong  \Delta CED

hence,

  • AE=CE
  • DE=BE

Proven

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