How to Customize index using code first?
GUID COMB strategy in EF
A graph $G$ is bipartite if and only if $G$ can be coloured with 2 colours Dump of my own Stack Exchange content Is there a way to quickly save all my Stack Exchange content across all Stack Exchange sites, as hundreds of HTML pages? (for me probably 15 or 20 sites: SO, dsp.SE, unix.SE, etc.)?  The ideal way would be one .HTML file for each question that I asked / answered / commented? What would be the ideal tool for that?    Remark: I've already read interesting questions like  but it doesn't exactly answer my concern here. What is the best way to generate a GUID/UUID from Apex code (such as a trigger)? Preferably in the following format:     nnnnnnnn-nnnn-nnnn-nnnnnnnnnnnnnnnnn   Example:     13219ec0-3a81-44c5-a300-de14b7d0235f Why can't I just think the spin as rotating? I'm going mad about the problem.  I really don't understand why do electron have 1/2 spin number, why they are not actually spinning.  I can accept that the electrons have their own magnetic field, which is certain, but why do they have $\hbar\sqrt3/2$ of angular momentum, and I don't know what the heck is spin number.  I've read the definition of  and  more than a hundred times but there is no betterment. I've smashed my head in my desk more than a hundred times either.  My question is the title. Why can't I just think the spin as rotating?  What I've saw recently,     electron's hypothetical surface would have to be moving faster than the speed of light in order for it to rotate quickly enough to produce the necessary angular momentum. How to use numpad / on a keyboard without a numpad? I've just got into Blender, and I like it a lot. But as I just switched from a stationary computer to a MacBook, I've got some new problems regarding the numpad. I can use 1-9 perfectly, But I have no idea how to emulate buttons like / (used to enter local view). I've searched the web for an answer, but I can't find any. I'm trying to find an alternative to the standard tombstone qed symbol for more informal papers. I initially thought of an empty square with Pub written inside, but what would actually be better would be a small beer mug symbol. (much like the one found in this set of icons )  Unfortunately there doesn't seem to such a symbol in the comprehensive latex symbol list (although many funny symbols can be found there). Has anyone thought about this? Is there any quick-and-dirty solution? Find $x,y$ given $\gcd(x,y)$ and ${\rm lcm}(x,y)$ Marginal probability derived from joint probability distribution of a poisson process and binomial process I thought I figured out this problem, but my answer is different than the solutions manual. Hoping someone here can stear me in the correct direction:     Assume the number of automobile collisions that occur on a given stretch of highway per year is a Poisson random variable $X$ with $\mu = 20$. The probability is $p=0.05$ that there will be one or more fatalities in each accident; occurrences of fatalities are independent from one collision to the next. If $Y$ is the number of collisions with one or more fatalities on this stretch of road in one year, find the probability law for $Y$.   So as said, X is poisson, so:  $$p_X(x) = \frac{20^x}{x!}e^{-20}$$  The conditional probability of $Y$ given $X=x$ is a Binomial random variable:  $$p_{Y\mid X}(y\mid x) = \binom{x}{y}p^yq^{x-y}$$  where $Y =$ number of fatal traffic accidents and $x =$ given number of accidents (from poisson distribution)  Therefore the joint probability for $X$ and $Y$ is:  $$p_{X,Y}(x,y)= p_{Y\mid X}\cdot p_X(x) = \binom{x}{y}p^yq^{x-y}\frac{20^x}{x!} e^{-20}$$  to find the marginal probability of $Y$, I sum over $P_{X,Y}$ for all $x$:  \begin{align} p_Y(y) &amp; = \sum_{x=y}^\infty{\binom{x}{y}p^yq^{x-y}\frac{20^x}{x!}e^{-20}} = \frac{(20p)^ye^{-20}}{y!} \sum_{x=y}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} \\[10pt] &amp; = \frac{(20p)^ye^{-20}}{y!} \sum_{x-y=0}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} = \frac{(20p)^ye^{-20}}{y!}e^{20q} \end{align}  So:   $$p_Y(y) = \frac{(20p)^ye^{-20p}}{y!}$$  however, the correct answer is apparently $p_Y(y)=\dfrac{e^{-1}}{y!}$ How can I copy a file from another directory to the current one? I have this problem as part of a school assignment:  NOTE: In case it's hard to read in the image above, here's the directory tree: $ tree . `-- sample_dir1     `-- sample_dir         |-- admin         |-- cambridge         |   |-- cafeteria         |   |-- library         |   `-- security         |       |-- annex         |       |-- building         |       `-- parking         |-- faculty         |-- history.exe         |-- markham         |   |-- annex         |   |-- building1         |   `-- parking         |-- oxford         |   |-- outline.doc         |   |-- programming         |   |   `-- report.pdf         |   `-- security         `-- stenton             |-- gen_ed             `-- lib_arts                 |-- english.txt                 `-- match.doc  15 directories, 11 files  I tried doing the command: $ cp ../cambridge/security/parking ./parking2  ...but it isn't working. The question in the image above states that the current directory is stenton, and that we need to make a copy of the file named parking (from the security directory), and to name the new file parking2, and place it in the current directory, stenton, using relative pathnames. What am I doing wrong? My cp command seems to be correct. This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. How can I figure out why my washing machine is tripping the GFCI receptacle? I have a Kenmore 80 Series (Model No. 110.24872300) washing machine which up to this point has been working fine. Since it is near a sink, the outlet it is plugged into is a GFCI outlet. On the last load I came back after letting it run for some time to find that it had tripped the GFCI in the middle of the wash cycle - the washer was still full of soapy water. I reset the outlet and let it run again, only to come down the next morning and find it full of water still (it had tripped during the rinse cycle this time). I reset the outlet again and this time it finished the rest of the cycle.  My guess is that water is somehow getting onto the internal components of the washer and causing it to short - is there any way I can test this theory or fix the problem myself? Is there a simple command for the available height in a beamer slide? Locally bounded additive function is linear Let $T:\mathbb{R}^k\to\mathbb{R}^m$ is additive, i.e. $T(X_1+X_2)=T(X_1)+T(X_2)$. $T$ is locally bounded. Prove that $T$ is a linear transformation.  I know there is a relation between boundedness, convexity and scalar multiplicity. But not sure where to start or how to tie things together. Which facility upgrade unlocks maneuver nodes in KSP Mafia 3 no more damage to be dealt barclay mills I did most of the barclay mills missions and brought the damage down to 2550 but now I can find anything else to damage or steal, so the meter is stuck at 2550 and the objective says "find and destroy garbage barges" but when I click it, no waypoint.  Please help. If I have random variables $X_1,X_2,\ldots,X_n$ that are Poisson distributed with parameters $\lambda_1, \lambda_2,\ldots, \lambda_n$, what is the distribution of $Y=\left\lfloor\frac{\sum_{i=1}^n X_i}{n}\right\rfloor$ (i.e. the integer floor of the average)?  A sum of Poissons is also Poisson, but I am not confident enough in statistics to determine if it is the same for the case above. Comaximal ideals in a commutative ring A question about the arctangent addition formula. In the arctangent formula, we have that:  $$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$  however, only for $uv&lt;1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?  One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:  $$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$  Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv&lt;1$ would mean $\tan\alpha\tan\beta&lt;1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2&lt;\alpha+\beta&lt;\pi/2$, i.e. that we have to stay in the same period of tangent.  However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation. How to install and use Ubuntu from a USB flash drive on Windows 8? Why do I need multiple sites for the same property in Google Webmaster Tools? In researching the best practice for adding sites to the Search Console, I've often seen it suggested to add separate properties for different versions of the same website. Even in  it states:     "Similarly, if your site has http:// and https:// versions, you should   add each as a separate site."   I've seen the same suggested for www/non-www. So if a site is accessible at all the variations of http/s and www/non-www we should have four properties.  But what if the site is only accessible at one variant? i.e. https and www are forced via htaccess, so only  can ever be accessed - should we still add all the variants?   In essence, the problem is that I don't know why Google wants us to do this, so I can't work out if it'll be ok just to have one instance.