Favorite color

Choose the correct level of measurement.

A) Interval

B) Nominal

C) Ratio

D) Ordinal

lwfrgin
2021-03-04
Answered

Determine the level of measurement of the variable.

Favorite color

Choose the correct level of measurement.

A) Interval

B) Nominal

C) Ratio

D) Ordinal

Favorite color

Choose the correct level of measurement.

A) Interval

B) Nominal

C) Ratio

D) Ordinal

You can still ask an expert for help

Yusuf Keller

Answered 2021-03-05
Author has **90** answers

Step 1

The level of measurement of the variable “favorite color” is nominal level of measurement, that is, option (B). The nominal level of measurement can be classified as measurement of data where numbers are only used to classify the data and numbers do not have any quantitative meaning. The words, letters and the alpha-numeric symbols can also be used in this level of measurement.

Step 2

The option, “Ordinal” can be correct because the favorite color does not have a defined order. Similarly, the properties of ratio and interval level of measurement is not fulfilled by the variable “favorite color”.

Hence, option (B), “Nominal” is the correct level of measurement.

The level of measurement of the variable “favorite color” is nominal level of measurement, that is, option (B). The nominal level of measurement can be classified as measurement of data where numbers are only used to classify the data and numbers do not have any quantitative meaning. The words, letters and the alpha-numeric symbols can also be used in this level of measurement.

Step 2

The option, “Ordinal” can be correct because the favorite color does not have a defined order. Similarly, the properties of ratio and interval level of measurement is not fulfilled by the variable “favorite color”.

Hence, option (B), “Nominal” is the correct level of measurement.

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The given equations are:

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The given equations are:

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C)

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Consider the set of natural numbers $\mathbb{N}$ with the $\sigma $-algebra $P(\mathbb{N})$ and the count measure $\mu $. If $f:\mathbb{N}\to [0,\mathrm{\infty})$, then

$\int fd\mu =\sum _{n=1}^{\mathrm{\infty}}f(n).$

I feel that I need to prove at first that f is a measurable function and then define a sequence ${f}_{n}\in {L}^{+}$ s. t. $f=\sum _{n}{f}_{n}$, but I have no idea if it's the right way nor how to do it. I need help.

Note: I can use any theorem, no restrictions.

$\int fd\mu =\sum _{n=1}^{\mathrm{\infty}}f(n).$

I feel that I need to prove at first that f is a measurable function and then define a sequence ${f}_{n}\in {L}^{+}$ s. t. $f=\sum _{n}{f}_{n}$, but I have no idea if it's the right way nor how to do it. I need help.

Note: I can use any theorem, no restrictions.

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A friend and I have been working on this problem for hours - how can the following expression be simplified analytically?

It equals $\frac{1}{2},$, and we have tried the following to no avail:

1. Substitution of $x=\sqrt{5}$

2. Substitution of $x=2\sqrt{5}$

3. Substitution of $x=5+\sqrt{5}$

4. Substitution of $x=\sqrt{5+\sqrt{5}}$

5. Manipulations by substituting the golden ratio

Here goes:

$\frac{{\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{2}}+{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}-{\displaystyle \frac{\sqrt{10+2\sqrt{5}}}{8}}}{{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}+5\cdot {\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{4}}}$

Thanks in advance for any help.

A friend and I have been working on this problem for hours - how can the following expression be simplified analytically?

It equals $\frac{1}{2},$, and we have tried the following to no avail:

1. Substitution of $x=\sqrt{5}$

2. Substitution of $x=2\sqrt{5}$

3. Substitution of $x=5+\sqrt{5}$

4. Substitution of $x=\sqrt{5+\sqrt{5}}$

5. Manipulations by substituting the golden ratio

Here goes:

$\frac{{\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{2}}+{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}-{\displaystyle \frac{\sqrt{10+2\sqrt{5}}}{8}}}{{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}+5\cdot {\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{4}}}$

Thanks in advance for any help.

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