Good night,  I've been studying a bit more about LaTeX, more specifically about TikZ and how to create images with it and I've come across this image on the manual:    The image above is on the 90th page of the manual (V 2.10).  The image above got my interest because I might teach statistics next semester and this could be a different way of showing data to my students but I didn't find it's code in the manual so I could study it (I'm afraid of having missed it though).  How can I draw something similar to that in TikZ? Thanks in advance.
I really like the 'ring chart' or 'wheel chart' given as an example on page 88 of the PGF manual. It is like a pie chart, but with a label in the centre. Unfortunately, the manual doesn't give the source for that chart. Does anyone have an example? Thanks.
Trying to plot the sin(x) function, I get this result:    The MWE is here:  \documentclass{standalone} \usepackage{ pgfplots} \begin{document} \begin{tikzpicture} \begin{axis} \addplot[samples=500]{sin(x)}; \end{axis} \end{tikzpicture} \end{document}   cos(x) looks like an inverted parabola and tan looks just like sin.  I'm using pdflatex and MiKTeX. Any ideas? How many ways can the average of n dice be a? How to rename objects with python? I have a  beautiful  inequality, but I can only prove part of cases. Given any positive real numbers $a,b,c$, we have   $$(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$$  How can we prove this statement? I'm struggling with the following question;     For every graph $G$ such that $K_3 \not\subseteq G$ (i.e. $G$ does not contain a triangle), prove that $\chi(G) \leq 2\sqrt{n} +1$ (where $\chi(G)$ denotes the chromatic number).   Any hints? I really cannot see the connection between the triangle-free nature of $G$ and a chromatic number being bounded above by $O(\sqrt{n})$. Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $ I need to find a reg ex that only allows alphanumeric. So far, everyone I try only works if the string is alphanumeric, meaning contains both a letter and a number. I just want one what would allow either and not require both. Changing unusual foreign currencies in London I've just returned from several unusual countries in Asia, and annoyingly despite being told I'd have no problem, I'm unable to find anyone in London to change back to pounds.  I've tried Travelex in Heathrow, a currency exchange place, and the post office.  The currencies include cash from: Mongolia, Kyrgyzstan, Kazakhstan and Uzbekistan.  Most of the responses when shown the money is to the tone of "What the hell currency is this?" which doesn't fill me with confidence.  Any suggestions welcomed for places that may change this. Otherwise I may need to go back next year ;) What else can you do in the turn you ready an action? MLE of $\theta$ when $X_1,\ldots,X_n $ are i.i.d with $P_\theta(x; \theta)=2x/\theta^2$ for $0\le x \le\theta$ Assume the standard situation, that is, let $X_1, \ldots , X_n$ be independent and identically distributed with $X_k ∼ P_\theta(x; \theta)$ , where $P_\theta(x; \theta) = 2x/\theta^2$  if $0\le x \le\theta$ and $0$ otherwise. It is required to estimate $\theta$. Show that the maximum likelihood estimator for $\theta$ is $\hat{\theta} = \max{[X_1, \ldots , X_n]}$ and then show that the cumulative distribution function of $\hat{\theta}$ is $F_\theta(z) = z^{2n}/\theta^{2n}$. Here's what i did so far: Maximum Likehood estimator: $L_x(θ) = \prod_{i = 1}^{n} P_θ(x_k)$ Here we have $P_θ(x_1,...,x_n;θ) = P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)$ Likehood = $L_{x,θ}(θ) =P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)= 2x_1/θ^2. 2x_2/θ^2.. 2x_n/θ^2 = [2^n.\prod_{i = 1}^{n}x_i]/θ^{2n}$ Log-likehood: $\sum_{i = 1}^{n}log(P_θ(x_1,...x_n;θ))= \sum_{i = 1}^{n}log(2x_i/θ^2)$ Is this correct so far? Im still not sure how to get to $\hat{θ} = max{[X_1, . . . , X_n]}$ As for the the cumulative distribution part to show $F_θ(z) = z^{2n}/θ^{2n}$: $F(z) = P(max({x_k})&lt;z) = P(x_1&lt;z).P(x_2&lt;z)..P(x_n&lt;z) = 2x_1/θ^2.2x_2/θ^2...2x_n/θ^n = 2^n.\prod_{i = 1}^{n}x_i/θ^2n$ Not sure if this is correct. Would really appreciate some help. Edit: From the answers below, we can deduce the the estimator is biased. What estimator would be unbiased? How can i find it? Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n  Hi everyone, for the proof to the above question,  Can I assume that since $(a, b) = 1$, then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the $nth$ power, they will still have no prime factors in common, and so $(a^n, b^n) = 1$ for all positive integers n.  I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :)  And also, I do not know how to approach the reverse problem where if $(a^n, b^n) = 1 then (a, b) = 1$, any guidance will be much appreciated! Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of Cardinal Arithmetic, which states that for $\kappa$ and $\lambda$ cardinal numbers, the larger infinite and the smaller nonzero, then $\kappa+\lambda=\kappa\cdot\lambda=\max(\kappa,\lambda)$.  I was playing around with the equation $k+\aleph_0=\mathfrak{c}$ for some cardinal $k$. From the above, it follows that $\mathfrak{c}=k+\aleph_0=\max(k,\aleph_0)$, which implies $k=\mathfrak{c}$.  I'm curious, can we still show $k=\mathfrak{c}$ without the Axiom of Choice? Is it maybe possible to bound $\mathfrak{c}-\aleph_0$ above and below by $\mathfrak{c}$? But then I'm not quite sure such algebraic manipulations even mean anything, or work like that here. Certainly normal arithmetic does not! Thanks. Why disable Fast Boot on Windows 8 when having dual booting? Why is everybody keep mentioning to disable Fast Boot on Windows 8 if you have installed alongside Ubuntu? Is it something that is recommended only for UEFI machines or is it a suggestion for Legacy BIOS machines as well? Is it because it makes the Windows partition inaccessible from Linux or there is another more serious reason to disable it? let $a,b,c,d,e$ are positive real numbers,show that $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$  My try:  I have solved follow Four-variable inequality:  let $a,b,c,d$ are positive real numbers,show that $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0$$ poof:since \begin{align*} &amp;\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\ &amp;=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4 \end{align*} By Cauchy-Schwarz inequality we have $$\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}$$ so we get $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0$$ Equality holds for $a=c$ and $b=d$  By done!  But for  Five-variable inequality,I can't prove it.Thank you Why didn't Skynet kill Kyle Reese? I have found film reels some with tails of film others not. Is there a way of telling if they have been exposed/used in a camera? Film types are for 35mm and one Kodak Advantix. Multi-window: prevent the out-of-focus app from pausing? How can I return to the start of a line in a console? How can I return to the start of a line and overwrite what has already been output on the console? The following does not appear to work:  System.out.print(mystuff+'\r'); In case, if I get a multiple-entry Schengen visa in one of the embassies in Ukraine, Kyiv - should my first trip be to the country which issued the visa?  Any there any requirements of this sort? Does this condition vary depending on embassy which issued the visa?  What are the possible consequences on not visiting that country first? Does it somehow depend on country where the embassy is (i.e. non-Schengen and not EU members)? How can I backup my RAW photos while travelling without Internet access? This question seems similar to , but my needs are different :   I shoot in RAW exclusively I'm looking for a 100% offline solution: while traveling, Internet won't be accessible the "laptop" solution is too heavy for me: I'm looking for a lightweight solution the simplest, the better: one device to rule them all   EDIT:  I'm using good old Compact Flash cards (yes, the bigger ones), so I need a device that can read it. If it can read from other cards, this question might be interesting for more people than myself  - that's the whole purpose of the website.    Also asked by :  Portable Storage Device while traveling  I need storage device which is able to read CF cards and make a backup of my photos while traveling. I do not want to carry laptop (it's bulky) and tablet (too little storage) but looking into some other solutions.   So far I found:   Epson PXXXX - price is blocker for me Jobo Giga Vu - same as above Nexto DI - bit better, but still a lot Hyperdrive Colorspace UDMA - this looks the best for price/options Wolwerine PicPac - cheap, bad opinions DigitalFoci PhotoSafe - same as above Ex-Pro® Picture 2 Drive - didn't find much about it   So ladys &amp; gents what do you use?