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3 years ago

A value of two standard deviation from the mean is more likely to occur than a value three standard deviations from the mean

Mathematics

2 answers:

ivolga24 [154]3 years ago

8 0

Answer:

true

Step-by-step explanation:

Anvisha [2.4K]3 years ago

3 0

This is true. The closer to the mean that a value is, the more likely it is to occur.

In this case, two and three standard deviations away are both fairly unlikely, however, two is still more likely to occur than three.

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Natasha2012 [34]

$\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]$

$\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]$

Argument of Complex number

Z=x+iy , is given by

If, x>0, y>0, Angle lies in first Quadrant.

If, x<0, y>0, Angle lies in Second Quadrant.

If, x<0, y<0, Angle lies in third Quadrant.

If, x>0, y<0, Angle lies in fourth Quadrant.

We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is

$\rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]$

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Nookie1986 [14]

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