trying to install add-on but not working
Fspy add-on not installing
how many conditional sentence types are possible? I wonder how many conditional sentence types are possible. Basically, there are four main and two mixed types. By mixing tenses, it could be up to nine of them. Are the rest viable?  I came across the following examples in Full Spate Pre-Profficiency Primer:   After you have just sold alcohol to a young guy. If he is if under eighteen, you should not have sold him the alcohol. If he promised to bring the drinks, he will. If you will take a seat, the doctor will see you shortly. (Likely, the former will demonstrates politeness only.) I would be grateful if you would tell me a little about your friends. (The same politeness, but would instead.) If he will keep eating all those biscuits, he can not expect to stay as slim as he used to be.   I would be grateful if someone could shed more light on the topic. Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See   For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.  In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$  My question is How can we differentiate between the two orbits? Does it have geometric meaning?    I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$... Let us denote by $\chi(G)$ the chromatic number, which is the smallest number of colours needed to colour the graph $G$ with $n$ vertices. Let $\bar{G}$ be the complement of $G$. Show that  (a) $\chi(G)+\chi(\bar{G}) \leq n+1$  (b) $\chi(G)\chi(\bar{G}) \geq n$  I was able to prove (a) using induction. Any hints on proving (b)? Calculating values of field based on two others using Field Calculator in ArcGIS Desktop? I need to combine two fields in an attribute table.  I want to do this using the field calculator.  What is the proper script to write to combine a field called "name" and "type", so that it produces one field with "name type"?  I'm using ArcGIS Desktop 10.0. Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ for $p\neq q$ distinct primes.  Where can I start with this proof? It looks similar to the Chinese Remainder Theorem, but that deals with two different a values. Does anyone know of a .NET date/time parser similar to  (handles stuff like "tomorrow" or "3pm next thursday")?  Note: I do write Ruby (which is how I know about Chronic) but this project must use .NET. Does a Readied spell immediately use a spell slot? When to use Stateful session bean over Stateless session bean? Show tabs in Chrome full screen mode? Firefox can do it, can Chrome?   Here's what I'm looking for: Show that $(p \to q) \lor (q \to p)$ is a tautology I tried to prove that $(p \to q) \lor (q \to p)$ is a tautology.  I used $p$ and $¬q$ as conditions. (Premises 1 and 5)  I managed to get to a solution, but I'm not sure if it's right.  Can you please check it?   Thank you!!!:) Covariance of Poisson and Conditional Binomial RV's Problem Statement  Let $X$ and $Y$ be random variables such that $X \sim \text{Poisson}(\lambda)$ and $Y|X \sim \text{Binomial}(x+1,p)$. Find $\text{Cov(X,Y)}$.  Attempt at a Solution  I would like to be able to write $Y$ as $X + Z$ where $Z$ is Poisson and independent of $X$, since then by  the variance is easy to compute. To this end, write $Y|X$ as a sum of independent binomials: $ Y|X = Z_1|X + Z_2|X, $ where  $Z_1|X \sim \text{Binomial}(1,p)=\text{Bernoulli}(p)$ and $Z_2 | X \sim \text{Binomial}(x,p).$ Then $Z_2 \sim \text{Poisson}(\lambda p)$, which we can obtain by directly computing the pdf: \begin{aligned} P(Z_2 = z) &amp;= \sum_x P(Z_2=z, X=x)\\ &amp;= \sum_x P(Z_2=z\mid X=x)P(X=x) \\ &amp;= \sum_{x=y}^\infty \frac{e^{-\lambda} \lambda^x}{x!}\frac{x!}{z!(x-z)!}p^z(1-p)^{x-z} \\ &amp;= \frac{(\lambda p)^z}{z!}e^{-\lambda p} \sum_{x=z}^\infty \frac{[\lambda(1-p)]^{x-z} e^{-\lambda(1-p)}}{(x-z)!} \\ &amp;= \frac{(\lambda p)^z}{z!}e^{-\lambda p}. \end{aligned} However, I cannot get the same approach to simplify for $Z_1|X$, which instead involves the CDF of $X$. I also tried this approach on $Y|X$ directly, hoping to obtain a sum in the form of the expectation of $X$, but I can't seem to reduce it. Even if I could, I'd still need to compute $E[XY]$ for the covariance. I suspect there may be a better approach than grinding through several more sums. 14 months ago, an  supplement was published that discussed how the ranger described in the Player's Handbook was under powered. These rules reimagined the ranger's abilities, and suggested that an official version would be released later.     Finally, we come to implementation. If this iteration of the ranger, or a future revision of it, grades high enough, our plan is to present it as a revised ranger in a future D&amp;D sourcebook.   Does that Sourcebook exist yet? If not, have the devs given an update or told us their intentions? Is a clean install better than upgrading? Is it better to install Ubuntu completely new or is the upgrade process just as good? In other words, will my computer run just as problem-free and efficiently when I upgrade as opposed to a fresh install? Suppose that $f$ is differentiable on $\mathbb{R}$ and $\lim_{x\to \infty}f'(x)=M$. Show that $\lim_{x\to \infty}f(x+1)-f(x)$ exists and find it. I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following: Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A &gt;0$  s.t. if $x&gt;A$ then $|f'(x)-M|&lt;\epsilon$. So we have that $M-\epsilon&lt;f'(x)&lt;M+\epsilon$. By MVT on $[x,x+1]$ for $x&gt;A$, $\exists c\in (x,x+1)$ s.t. ${f(x+1)-f(x)\over 1}=f'(c)$. Therefore, $M-\epsilon&lt;f(x+1)-f(x)&lt;M+\epsilon$. However, I do not know what to do after this. I know that there is a simple way to shutdown the computer in command prompt, but is it possible to do system updates and then shutdown the computer in command prompt?   For example, when Vista needs to install updates, it will let me install them and shut down by clicking the Shut Down button (while relegating the normal shut down button into the menu). So is there a command to do this in command prompt? Do we have more history for cd? Evaluate $\lim_{y\to 0} \dfrac {(x+y)\sec (x+y) - x\sec x}{y}$ Evaluate $\lim_{y\to 0} \dfrac {(x+y)\sec (x+y) - x\sec x}{y}$.  My Attempt: $$=\lim_{y\to 0} \dfrac {(x+y) \sec (x+y) - x\sec x}{y}$$ $$=\lim_{y\to 0} \dfrac {x\sec (x+y) + y\sec (x+y) - x\sec x}{y}$$ $$=\lim_{y\to 0} \dfrac {x(\sec (x+y) - \sec x) + y\sec (x+y)}{y}$$ Wubi with 14.10 doesn't work - failed to partition the selected disk I tried to install Ubuntu 14.10 on my computer with Wubi. The installation appeared to work in Windows, and told me to reboot.   After rebooting the Ubuntu installer continued the installation, until it stopped with an error message: Failed to partition the selected disk with the detail This probably happened because are too many primary partitions in the partition table. If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if   $\bullet \ A=\emptyset, \ B\neq\emptyset$,?  $ \bullet \ B=\emptyset, \ A\neq\emptyset$ ?  $ \bullet \ B=\emptyset, \ A=\emptyset$ ?  (Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? Are they all $\emptyset$ ?) Why did Aragorn and Theoden ride out at the Hornburg? In the Battle of the Hornburg from The Two Towers, Aragorn and Theoden say "We must ride out." Why did they do this?  Was it just a matter of honor? Did they ride out on their horses in the middle of battle because they thought it was the end, as they were outnumbered? Was it because Aragorn knew Gandalf was coming so they were confident they were about to win?