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Let G be a finite group with ${\displaystyle card\left(G\right)=p^{2}q}$ with ${\displaystyle p<q}$ two ' numbers. We denote ${\displaystyle s_q}$ the number of q-Sylow subgroups of G and similarly for p. I have just shown that $s_q\in \{1, p^2\}$. Now I want to show that
$\underset{S\in Sy{l}_q\left(G\right)}{\cup }S\setminus \left\{1\right\}={\dot{\cup }}_{S\in Sy{l}_q\left(G\right)}S\setminus \left\{1\right\}$
i.e. that for ${\displaystyle S,T\in Sy{l}_q\left(G\right)}$ with ${\displaystyle S\ne T}$ we that $\mathrm{S}\setminus \left\{1\right\}\cap \mathrm{T}\setminus \left\{1\right\}=\varnothing$

Given you have an independent random sample ${\displaystyle X_1,X_2,\dots ,X_n}$ of a Bernoulli random variable with parameter $p$, estimate the variance of the maximum likelihood estimator of $p$ using the Cramer-Rao lower bound for the variance
So, with large enough sample size, I know the population mean of the estimator ${\displaystyle \hat{P}}$ will be $p$, and the variance will be:
$Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}$
Now I'm having some trouble calculating the variance of ${\displaystyle \hat{P}}$, this is what I have so far:
since the probability function of ${\displaystyle \bar{X}}$ is binomial, we have:
$f_x\left(\overline{X}\right)=\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)*p^{\sum _{i=1}^n{X}_i}*(1-p{)}^{n-\sum _{i=1}^n{X}_i}$
so: $\ln \hspace{0.17em}{f}_X\left(X\right)=\ln \left(\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)\right)+\sum _{i=1}^n{X}_i\ln \left(p\right)+\hspace{0.17em}(n-\sum _{i=1}^n{X}_i)\ln (1-p)$
and: $\frac{\partial \ln \hspace{0.17em}{f}_X\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^n{X}_i}{p}-\frac{(n-{\sum }_{i=1}^n{X}_i)}{(1-p)}=\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}$
 

and: $(\frac{\partial ln\hspace{0.17em}{f}_X\left(X\right)}{\partial p}{)}^2=(\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}{)}^2=\frac{n^2{p}^2-2n^{2}p\overline{X}+n^2{\overline{X}}^2}{p^2(1-p{)}^2}$
since ${\displaystyle E\left[{\bar{X}}^2\right]={\mu }^2+\frac{{\sigma }^2}{n}}$, and for a Bernoulli random variable ${\displaystyle E\left[X\right]=\mu =p=E\left[\bar{X}\right]}$ and ${\displaystyle Var\left[X\right]={\sigma }^2=p(1-p)}$:
$E\left[(\frac{\partial \ln \hspace{0.17em}{f}_X\left(X\right)}{\partial p}{)}^2\right]=\frac{n^2{p}^2-2n^{2}pE\left[\overline{X}\right]+n^{2}E\left[{\overline{X}}^2\right]}{p^2(1-p{)}^2}=\frac{n^2{p}^2-2n^2{p}^2+n^2(p^2+\frac{p(1-p)}{n})}{p^2(1-p{)}^2}=\frac{np(1-p)}{p^2(1-p{)}^2}=\frac{n}{p(1-p)}$
Therefore, $Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}=\frac{1}{n\frac{n}{p(1-p)}}=\frac{p(1-p)}{n^2}$
However, I believe the true value I should have come up with is ${\displaystyle \frac{p(1-p)}{n}}$.

Find the area of the largest that can be inscribed in a semicircle of radius r.

the manager of an electrical supply store measured th diameters of the rolls of wire in the inventory. The diameter of the rolls are listed below 0.56,  0.622, 0.154, 0.412, 0.287, 0.118

find an equation in rectangular coordinates p+3=e and r-2 = 3cosΘ