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2 years ago

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As an introduction to probability, a student asked to roll a fair, six sided number cube seven times the results of those seven

rolls are shown below 1,4,4,4,4,6,5 what is the standard deviation of the data?

1 answer:

frosja888 [35]2 years ago

3 0

Using it's concept, the standard deviation of the data is of 3.742.

<h3>What are the mean and the standard deviation of a data-set?</h3>

The mean of a data-set is given by the <u>sum of all values in the data-set, divided by the number of values</u>.

The standard deviation of a data-set is given by the square root of the <u>sum of the differences squared between each observation and the mean, divided by the number of values</u>.

For this data-set, the mean is:

M = (1 + 4 + 4 + 4 + 4 + 6 + 5)/7 = 4.

Hence the standard deviation is:

$S = \sqrt{\frac{(1 - 4)^2 + (4 - 4)^2 + (4 - 4)^2 + (4 - 4)^2 + (4 - 4)^2 + (6 - 4)^2 + (5 - 4)^2}{7}} = 3.742$

More can be learned about the standard deviation of a data-set at brainly.com/question/12180602

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3 years ago

What is the completely factored form of x3-64?

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Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
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We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

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