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vodka [1.7K]
3 years ago
11

Help me find Sample variance and standard deviation

Mathematics
1 answer:
AnnZ [28]3 years ago
5 0
First, find the mean. You'll need it to compute the variance.

\bar x=\displaystyle\frac15\sum_{i=1}^5x_i=\dfrac{21+10+6+7+11}5=11

The variance of the sample is then computed with the formula

s^2=\displaystyle\dfrac1{5-1}\sum_{i=1}^5(\bar x-x_i)^2=\dfrac{(21-11)^2+\cdots+(11-11)^2}4=\dfrac{71}2=35.5

The standard deviation is the square root of the variance, so

s=\sqrt{35.5}\approx5.958
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Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.
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<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with f^{-1}(x)

Given that:

I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = a^{-1}(x) = a(x)  Hence, I is true.

II. y =\dfrac{1}{x}

Replacing y with and x with y:

x =\dfrac{1}{y}

x=\dfrac{1}{a^{-1}(x)}

\Rightarrow a^{-1}(x) = \dfrac{1}{x}

a^{-1}(x) = a(x)  Hence, II is true.

III. y =x^{2}

Replacing y with and x with y:

x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)

 Hence, III is not true.

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Replacing y with and x with y:

x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)

Hence, IV is not true.

<em>Correct answer:</em>

<em>A. I and II</em>

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