Please excuse my noobs query, as I'm new to linux environment. I'm getting errors in sudo apt-get update  W: Failed to fetch http://deb.playonlinux.com/dists/precise/InRelease  The following signatures couldn't be verified because the public key is not available: NO_PUBKEY E0F72778C4676186 E: Failed to fetch http://ppa.launchpad.net/pmcenery/ppa/ubuntu/dists/xenial/main/source/Sources  404  Not Found W: Some index files failed to download. They have been ignored, or old ones used instead.   There was some error in this ppa.launchpad.net/pmcenery, I found some suggestion after googling, it said me to remove the key by doing  sudo apt-get del Keyno. I did removed that still there are some errors, now it is saying there's no public key.  Please suggest how to get rid of this issue.
I added some extra repositories with the Software Sources program. But when I reload the package database, I get an error like the following:     W: GPG error:  trusty InRelease: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY 8BAF9A6F   I know I can fix it using apt-key in a terminal, according to the official Ubuntu documentation. But I would have liked to do it graphically. Is there a way to do this without using a terminal?
Dual boot Ubuntu (14.04) with Windows 10 Probability Number A greater Number B Given $a \in \{1,2,...,250\}$ and $b\in \{0,1,...,1000\}$  What is the probability that $a &gt; b$?  I know there already is  question with a very similar question, however it's over an interval of $\mathbb{R}$ so I hope there is another way opposed to the Integral used in the formal solution.  How does one go about calculating said probability? I have a TikZ picture with a node containing a minipage with an itemize environment; each item is a hyperref to other sections of the document.   When scaling the image (for example with \begin{tikzpicture}[scale=.7, transform shape] I've a strange behavior: links in itemize are not in the right place: they present a somewhat vertical offset, as if they don't scale as the rest of image.   Here the MWE:  \documentclass[a4paper]{article} \usepackage{tikz} \usepackage[pdftex]{hyperref} \begin{document}  Here something interesting\label{text:interesting-one} \newpage  Here something interesting, too\label{text:interesting-two} \newpage  and here there are very very interesting stuff\label{text:interesting-three} \newpage  \newpage Now the problem: links rectangles have some offset, making links unusable in case of long itemize  \begin{figure}[htb] \begin{tikzpicture}[scale=.7, transform shape] \node at (0,0) (mynode) { \begin{minipage}{4cm} \begin{itemize} \item \hyperref[text:interesting-one]{First link} \item \hyperref[text:interesting-two]{Second link} \item \hyperref[text:interesting-three]{Third link} \item \hyperref[text:interesting-one]{First link again} \item \hyperref[text:interesting-two]{Second link again} \item \hyperref[text:interesting-three]{Third link again} \end{itemize} \end{minipage} }; \end{tikzpicture} \end{figure} \end{document}   And here a snapshot of the problem    Note: This is evident using the default settings of hyperref package, which draws the red rectangle around links, but it took me some time to found it when only the text is coloured, i.e. when using \usepackage[linkcolor=blue,colorlinks=true]{hyperref} What is the typical orbital life of an artificial satellite? The orbit of satellites around Earth eventually decays, or so I read. This is typically caused either by atmospheric drag, or by tides. I would assume most satellites have a limited service life in orbit. Hence the question - What is the typical orbital life of a satellite? How long before it's successor must be launched into orbit ?  EDIT: For instance, a weather satellite Watched screen recording but it's not saved to the desktop I recently recorded a session with my dad. I stopped the recording after the session ended (with the setting set to save to Desktop), clicked the icon that appeared on the lower right corner to open the file, and played back a few minutes to make sure it worked. I then clicked Share to send it via AirDrop to my computer, which did not work; the status remained &quot;cancelled&quot; despite my efforts. I assumed I'd be able to access it through my phone by sending it through a third app, so I pressed cancel on the AirDrop pop-up window. The entire screen recording closed and (apparently) vanished forever after the action. I'm extremely frustrated because I can't seem to locate this video. I am trying to prove the statement:  If $X\sim\mathcal{N}(0,\sigma_1^2)$ and $Y\sim\mathcal{N}(0,\sigma_2^2)$ are independent random variables, then $\frac{XY}{\sqrt{X^2+Y^2}}$ is also a Normal random variable.  For the special case $\sigma_1=\sigma_2=\sigma$ (say), we have the well-known result that $\frac{XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ whenever $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables. In fact, it is more generally known that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ variables. A proof of the last result follows by using the transformation $(X,Y)\to(R,\Theta)\to(U,V)$ where $x=r\cos\theta,y=r\sin\theta$ and $u=\frac{r}{2}\sin(2\theta),v=\frac{r}{2}\cos(2\theta)$. Indeed, here $U=\frac{XY}{\sqrt{X^2+Y^2}}$ and $V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$. I tried to imitate this proof for the problem at hand but it appears to get messy. If I haven't made any error, then for $(u,v)\in\mathbb{R}^2$ I end up with the joint density of $(U,V)$ as $$f_{U,V}(u,v)=\frac{2}{\sigma_1\sigma_2\pi}\exp\left[-\sqrt{u^2+v^2}\left(\frac{\sqrt{u^2+v^2}+v}{\sigma_1^2}+\frac{\sqrt{u^2+v^2}-v}{\sigma_2^2}\right)\right]$$ I have the multiplier $2$ above as the transformation is not one-to-one. So density of $U$ would be given by $\displaystyle \int_{\mathbb{R}}f_{U,V}(u,v)\,\mathrm{d}v$, which isn't readily evaluated. Now I am interested to know if there is a proof where I can only work with $U$ and don't have to consider some $V$ to show that $U$ is Normal. Finding the CDF of $U$ doesn't look so promising to me at the moment. I would also like to do the same for the case $\sigma_1=\sigma_2=\sigma$. That is, if $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables then I wish to show that $Z=\frac{2XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}(0,\sigma^2)$ without using a change of variables. If somehow I can argue that $Z\stackrel{d}{=}X$, then I'm done. So two questions here, the general case and then the particular case. Related posts on Math.S.E: . . Edit. This problem is in fact due to L. Shepp as I found out in the exercises of An Introduction to Probability Theory and Its Applications (Vol. II) by Feller, alongwith a possible hint:  Surely, $U=\frac{XY}{\sqrt{X^2+Y^2}}=\frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}}$ and I have the density of $\frac{1}{X^2}$ at hand. Let's see what I could do now. Apart from this, a little help with the integral above is also welcome. Setup SunFire 440 Raid Configuration vertical alignment of text in tikz node The output from the following script \documentclass{article} \usepackage{tikz} \usepackage{graphicx} \begin{document} \begin{tikzpicture}   \node(title)[text width = 0.25\textwidth, text height = 3 cm, draw]{title};   \end{tikzpicture} \end{document}  is  How can I vertically align the text in the node? Specifying the text depth as in \node(title)[text width = 0.25\textwidth, text height = 1cm, text depth = 1 cm, draw]{     title line 1\\     title line 2\\     title line 3};  Does not result in the text being vertically centred. ICANN charges 18¢ per domain name registered regardless of its extension, but I pay $10 for a domain name per year with my registrar.   What do domain name registrars do for its customers to justify this extra $9.82?  Is there something that they do for our domain names other than registering it with ICANN?   I just read about the 7 people who hold the ICANN database key and got a little bit curious to know what the domain name servers do for us to justify this extra amount (12 times the actual fee)? Change the height of a function being plotted What is the best way to "carve" a terrain created from a heightmap? I have a 3d landscape created from a heightmap. I'd like to "carve" some holes in that terrain. That will allow me to create bridges, caverns and tunnels inside it.    That operation will be done in the game editor so it doesn't need to be realtime. In the end, rendering is done using traditional polygons.  What would be the best/easiest way to do that ? I already think about several solutions :  Solution 1  1) Create voxels from the heightmap (very easy). In other words, fill a 3D array like this : voxels[32][32][32] from the heightmap values.  2) Carve holes in the voxels as i want (easy too).  3) Convert voxels to polygons using some iso-surface extraction technique (like marching cubes).  4) Reduce (decimate) polygons created in 3).  This technique seems to be the most promising for giving good results (untested). However the problem with marching cubes is that they tends to produce lots of polygons thus reducing them is mandatory. Implementing 4) also seems not trivial, i have read several papers on the web and it seems pretty complex. I was also unable to find an example, code snippet or something to start writing an algorithm for triangle mesh decimation. Maybe there is a special decimation algorithm (simpler) for meshes created from marching cubes ?  Solution 2  1) Create some triangle mesh from the heighmap (easy).  2) Apply severals 3D boolean operation (eg: subtraction with a sphere) to carve the mesh.  3) apply some procedure to reduce polygons (optional).  Operation 2) seems to be very complex and to be honest i have no idea how to do that. Also applying many boolean operation seems to be slow and will maybe degrade the triangle mesh  every time a boolean operation is applied. LINQ - Query syntax vs method chains & lambda Frattini subgroup $Φ ( G )$ is not trivial Frattini subgroup $Φ ( G )$ of a group G is the intersection of all maximal subgroups of $G$. The question is  Let $G$ be an abelian group.What is the sufficient and necessary condition of $Φ ( G ) ≠｛e｝$？  I think if $G$ has no torsion part, we cannot say $Φ ( G ) ≠｛e｝$,because $\Bbb Z ×\Bbb Z×・・・ ×\Bbb Z$ has a maximal subgroup $p\Bbb Z ×\Bbb Z×・・・×\Bbb Z$, with intersectioning distinct primes, we gain just a trivial group. So, we need torsion part. But I cannot proceed from here.I only know the answer, $G$ has an element of order $p^2$, that  is, $G$ has theree subgroups. I want to deduce this consequence from steady discussion. P.S  According to this answer, in a ring that is a finite direct product of quotient rings of $\mathbf{Z}$, we can say rattini subgroup is the same as the Jacobson radical of the corresponding ring. HTTP request parameters are not available by request.getAttribute() Did she judge him "wrong" or "wrongly"? Which one is the correct use?  She judged him wrong. She judged him wrongly.  Or, are both correct, but have slightly different meanings? In PHP 5, what is the difference between using self and $this?  When is each appropriate? QGIS Align Rasters how to retain pixel values? I am attempting to align and then merge a number of rasters of the same resolution and extent, but where the pixels are slightly misaligned (creating small gaps when merged). This works fine using the Align Rasters tool, except that the pixel values are not retained and instead given a value of something like -3.23879e+8. I mostly want to keep the nodata pixels. Is this possible? When I studied Bayesian statistics, a question about the notation of Bayes' Theorem came to my mind. Below is the density function version of Bayes' Theorem, where $y$ is data vector and $\theta$ is the parameter vector:  $$ p(\theta|y)=\frac{p(y|\theta)p(\theta)}{p(y)} $$ The numerator on the right handside can be written as: $$ p(y,\theta) $$ which is the joint probability distribution of $y$ and $\theta$, then Bayes' theorem could be written as: $$ p(\theta|y)=\frac{p(y,\theta)}{p(y)} $$ Furthermore, $$  p(\theta|y)\propto p(y,\theta) $$ Am I right on this? I think it does not look right. Because the posterior is the proportional to the joint density function. But where is the mistake? How can I add "Show desktop" to the GNOME dash or Ubuntu Dock? How to check whether a string contains a substring in JavaScript?