Your sample is normally distributed with a mean age of 36. The standard deviation in this sample is 4 years. You would expect:

Mathematics

1 answer:

Nostrana [21]2 years ago

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Answer:

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Step-by-step explanation:

Given a normal distribution with ;

Mean = 36

Standard deviation = 4

According to the empirical rule :

68% of the distribution is within 1 standard deviation of the mean ;

That is ; mean ± 1(standard deviation)

68% of subjects :

36 ± 1(4) :

36 - 4 or 36 + 4

Between 32 and 40

2.)

95% of the distribution is within 2 standard deviations of the mean ;

That is ; mean ± 2(standard deviation)

95% of subjects :

36 ± 2(4) :

36 - 8 or 36 + 8

Between 28 and 44

3.)

99% is about 3 standard deviations of the mean :

That is ; mean ± 3(standard deviation)

99% of subjects :

36 ± 3(4) :

36 - 12 or 36 + 12

Between 24 and 48

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Which is equivalent to √60n^11/256n^4 after it has been simplified completely? Assume n =0

motikmotik

Answer:

$\large\boxed{\dfrac{n^3}{8}\sqrt{15n}}$

Step-by-step explanation:

$\sqrt{\dfrac{60n^{11}}{256n^4}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{60}{256}n^{11-4}}=\dfrac{\sqrt{60n^7}}{\sqrt{256}}=\dfrac{\sqrt{4n^{6+1}\cdot15}}{16}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{\sqrt{4n^6\cdot15n}}{16}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt{4n^{3\cdot2}}\cdot\sqrt{15n}}{16}\qquad\text{use}\ (a^n)^m=a^{nm}$

$=\dfrac{\sqrt{4(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{\sqrt{4}\cdot\sqrt{(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{2\!\!\!\!\diagup^1\sqrt{15n}}{16\!\!\!\!\!\diagup_8}=\dfrac{n^3\sqrt{15n}}{8}$