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 #4
avatar+2489 
+2

Hi Melody,

You correctly solved this problem here: https://web2.0calc.com/questions/probability-question_88#r3

 

Confirmation: \(\large (2^3) / (nCr(6,3) =  \dfrac {2}{5} = 40\%\)

 

 

GA

--. .-

Feb 13, 2022
 #9
avatar+118658 
+1

Thanks Tiggsy

I can follow all that.  Similar to mine just better :))

 

\(\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{24 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25 \)

 

 

 

 

 

 

LaTex

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)}
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)}
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{24
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25

Feb 13, 2022
 #1
avatar+118658 
+1
Feb 13, 2022
 #1
avatar+118658 
+1

Do you know what y=cos(x) looks like?

Graph it on desmos if you do not know..     

What is its period?

Where does it first cross the x axis?

https://www.desmos.com/calculator/xbk6uv697u

 

Try drawing cos(12x)   how does it compare?

Try drawing them, discuss it here, if you have specific questions then ask with a new post on this thread.

Feb 13, 2022
 #1
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Feb 13, 2022

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