mysql_install_db on Ubuntu 16.04
How to create a database directory structure for MySQL?
I need to calculate the time complexity of the following code:  for (i = 1; i &lt;= n; i++) {   for(j = 1; j &lt;= i; j++)   {    // Some code   } }   Is it O(n^2)? Remove duplicate fields in a given column (after ID check) How do I remap multi-key sequences to individual keys? I just bought an IBM Model M keyboard. While awesome, it lacks a windows key and multimedia keys. While there are many programs out there that can map one key to another, or one key to a key sequence, are there any programs that will map a keyboard sequence to a single key?  Edit: OS = Win7 Is $|xy|$ differentiable in $(0,0)$? Is $f(x,y) = |xy|$ differentiable in $(0,0)$? I have no idea how to approach this problem. The search parameter isanswered ignores acceptances Find the probability of an event determined by an inequality on exponential random variables Let X and Y be independent random variables such that $X∼Exp(1)$ and $Y∼Exp(2)$.  Find the probability that $3X+4Y≤5$.  I thought the equation for solving this was the integral from $\int_0^a \int_ 0^{(a-y)} \lambda_1 e^{-\lambda_1 x}\lambda_2 e^{-\lambda_2 y} dx dy$ where $a = 5$, so I first started with the integral from $\int_0^5 \int_0^{(5-y)} 3 \times 1e^{-1x} \times 4 \times 2e^{-2y} dx dy$ and got $23.83828927$ which is much larger than a probability can be. Any help? How do I enable comments for user profiles? With users being entities in D7 I assumed the user type would would have built in "comment" option. That doesn't seem to be the case. I just want to allow users to leave comments on the user profile page. Is there a simple way to doing that? Please help me to find elements $a,b$ in a ring $R$ such that $a\mid b$ and $b\mid a$, but there does not exist any unit $u$ in $R$ such that $a=ub$. Bounded sequence which is not convergent, but differences of consecutive terms converge to zero I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.  What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?  By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not. Ubuntu stuck/freeze on install I've been trying to replace my current Windows 10 installation with Ubuntu 16.04. My computer is a Lenovo Thinkpad T470. I've formatted my 15GB USB-dongle. Downloaded the software Rufus, and burned the ISO onto the USB. I've tried the settings "MBR Partition for BIOS or UEFI", "MBR Partition for UEFI" and "GPT partition for UEFI". I've tried using memtest86+ to test if I have any memory errors, I have not. I've tried disabling Secure boot, and Fast startup. I've tried using other software than Rufus, and I've tried entering the "nomodeset" in the boot options, as well "acpi=off", nothing works.   When I boot from my USB. I choose "Install Ubuntu" from the grub menu. And then select my language in the installer. It is from this point on, it gets problematic. Sometimes when I try to connect to my WIFI, it connects successfully, but when I press the "continue" button, the installer freezes and nothing happens. Sometimes I get past this point, but it then freezes later on in the installation progress - typically right when I press the "Install now" button.   It should also be mentioned that, if I press the "Quit" button in the install. I get to a black screen with a lot of text saying something like squashfs error.   Can someone please help me? When I tried booting memtest86+ it only worked when I changed my boot mode from UEFI to Legacy. Could this be it? Are there any downsides to installing an operating system in legacy?   If anyone has any other suggestions, it would be greatly appreciated!  Thank you :-) Why does my material panel look different? Complete beginner here.  I'm trying to follow  tutorial on Anisotropic shaders, but my property panel looks nothing like the property panel in the tutorial.  For example, when creating a material, this is what his property panel looks like:    But this is what mine looks like, if I follow the same path:    I'm sure I'm missing something obvious.  How can I see the property panel shown in the tutorial? I have a population density raster and would like to create 5 polygons from the 5 natural break (jenks) classification ranges. 1 polygon for each classification range. I'm using ArcMap10 and have all of the Extensions enabled in the ToolBox to my disposal.   Thanks! Conditional Expectation of Composite Function Preliminaries  Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.  Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).  Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.  Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.  Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.  Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.  Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi)  $$  For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.  Question  I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by  $$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)  $$ whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$  How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.  Thanks for any advice! How force Chrome to use google.com search engine without https Chrome's default search engine is .  How we can force chrome to use  without https?  I need a solution that allows proxies to rewrite my URLs, but https currently makes this impossible... Why does the CAN bus use a 120 ohm resistor as the terminating resistor and not any other value? Most instructions for equipment and software make a habit of shortening sentences by dropping words of certain parts of speech - usually definite and indefinite articles.  For example:     PULL (THE) LEVER TO OPERATE      CARS MAY BE TOWED WHEN (THEY ARE) PARKED ILLEGALLY      PLEASE USE (THE) OTHER DOOR   Is there a name for this custom? And what is the best way to describe it? Sometimes when reading through user comments on a particular answer, some usernames that appear like user[digit] e.g user12345 are not links. Is this a known behaviour or any other thing I am not aware of? What is involved in converting a PC video card to have boot screen on the Mac? It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.  The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.  This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m&gt;n$? This does not seem obvious.  Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get  $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.   The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$? Need to draw a flowchart as in the image below with some arrows I have problem with the feed back from bottom block to on top of top block. Can anyone help?