Hint: To generate this number you could perform the experiment of tossing fair coin 10 times.    I guess we are supposed to use the normal approximation to binomial distribution. But I'm stuck after that. Can anyone help me?
How can I manually generate a random number from a given distribution, as for instance, 10 realisations from the standard normal distribution?
Induction Proof: Fibonacci Numbers Identity with Sum of Two Squares Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step.  Identity: $$f_n^2 + f_{n+1}^2 = f_{2n+1}$$  I get to:  $$f_{n+1}^2 + f_{n+2}^2$$  Should I replace $f_{n+2}$ using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step?  Thanks! Irreducible polynomials and affine variety Let $k$ be any field, and let $f,g\in k[x,y]$ be two irreducible polynomials such that $g$ is not divisible by $f$. Prove that $V(f,g)\subseteq A_k^2$ is finite. If I select a feature and run self.iface.actionIdentify().trigger() it will open the identify window but won't open the selected feature's data, how can I make it open the data of the selected feature? Parse PHP code to extract function names? Is there a better way than simply trying to open the file?  int exists(const char *fname) {     FILE *file;     if ((file = fopen(fname, "r")))     {         fclose(file);         return 1;     }     return 0; } Trouble understanding Landau & Lifshitz writing about Lagrangians and Galilean Relativity We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$  $"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$  I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative? Show $GCD(a_1, a_2, a_3, \ldots , a_n)$ is the least positive integer that can be expressed in the form $a_1x_1+a_2x_2+ \ldots +a_nx_n$ Is it possible to prevent LUDecomposition from pivoting? Suppose that $U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$,  where $X_1,X_2,\ldots$ and $Y_i,Y_2, \ldots$ are i.i.d. sequences of random variables with mean $\mu$ and variance $\sigma^2$ and the sequences $\{X_i\}$ and $\{Y_i\}$ are independent.  Then, for all real numbers $x$, show that, as $n\to \infty$,  $$\mathbb{P}(U_n\leq x) \to  \int_{-\infty}^{x} \frac{1}{(\sqrt(2\pi))} \:e^{-\frac 12{u^2}}\mathrm du $$ Approach  I tried to carry out the integration on the right, but this does not integrate nicely. It ends up involving an error function, so I figured this was the wrong was to go.  Should I be looking at convergence in probability to answer this question? Is Chebyshev's inequality relevant here? If so, how and how does that relate to the integral? Suppose someone from "Harry's world" married a Muggle, and they want to escort their child. Can the Muggle parent get through to the Hogwarts platform? Why are news headlines in present tense if they refer to past events? In news papers we see headings like this  &quot;India signs a pact with Russia&quot; &quot;Sachin hits another century&quot; &quot;Obama wins presidential election&quot;  These are completed events, aren't they? Then, why are these sentences not mentioned in past tense? Why can't I tame my ocelot? I tried to change network interfaces name on this new Ubuntu 16.04 LTS version but doesn't have the /etc/udev/rules.d/70-persistent-net.rules.  So, I tried to use /lib/udev/write_net_rules but it not exist.  Why do I need this modification? Because I'm using a tool to Simulate virtual platforms that use flex licensing and the authentication need to be in eth0 interface name.  Any suggestion?  The command ip link returns:  user@laptop:~$ ip link  1: lo: &lt;LOOPBACK,UP,LOWER_UP&gt; mtu 65536 qdisc noqueue state UNKNOWN mode DEFAULT group default qlen 1 link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00 2: enp6s0: &lt;NO-CARRIER,BROADCAST,MULTICAST,UP&gt; mtu 1500 qdisc pfifo_fast state DOWN mode DEFAULT group default qlen 1000 link/ether b8:2a:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff 3: wlp7s0: &lt;BROADCAST,MULTICAST,UP,LOWER_UP&gt; mtu 1500 qdisc mq state UP mode DORMANT group default qlen 1000 link/ether 5c:e0:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff  user@laptop:~$ ifconfig  enp6s0    Link encap:Ethernet  HWaddr b8:2a:xx:yy:xx:yy   lo        Link encap:Local Loopback   wlp7s0    Link encap:Ethernet  HWaddr 5c:e0:xx:yy:xx:yy     (Some information were ignored and suppressed)  I already tried some links: Are the suggested edit audits too easy? Yes, this might be another post complaining about audits. But, in this case, I want the audits to be more difficult.  Specifically, I'm concerned that the suggested edits audits are just too darn easy to spot. They don't behave anything like actual edits, and so after a tiny bit of experience (and electric shocks from audit failures) I think they wouldn't slow down a roboreviewer at all.  Unlike the other queues, which use real data (and are legitimately, in my opinion, difficult), the suggested edits audits look like this:    In about 500ms I can tell that is a bogus edit (less time than it takes for the buttons to unfreeze). After just a few of these, I'm about 99% confident I could tell you which edits were audit edits and which ones were not, and I don't think I actually have to spend extra time on them. On the other queues, picking out the audits is very hard because they use real data, and so I wouldn't bother to do so.  Were I a roboreviewer, I could get on with my Accept-happy life, only pausing to reconsider when someone actually changes the text. (Seriously -- how many real edits have you see which change the text in such a drastic manner? 95% of edits are grammar, spelling, formatting or some combination of those).  Are the suggested edits audits too easy? Is the "Markov-chain random-edit-word-salad" too easily detectable? Or does it legitimately slow down and/or deter robo-reviewers? Should we (and can we) be using real data for the suggested audits queue too? I'm a web designer and I'd like to make a custom wordpress theme for my upcoming blog. The template is already made in photoshop but I need a good starting point. I don't have too much experience with Wordpress but I want something that is SEO friendly which I can edit without risking getting poor search results.  I Know about Starkers and that there are other blank themes out there, but are they good for SEO? I've often wondered why Stack Exchange has opted to keep the upvote/downvote buttons available for clicking on question/answers posted by the user. For instance:    Even though these buttons have no use for the user, why are they still placed on the post's left hand column? Moreover, the comments have no controls for flagging/+1'ing the user's own comment. What is the reason for this discrepancy? What benefit do the +1/-1 buttons provide to the user, even though they can't be used for anything? The question is:      Prove that if $p \in \mathbb{Z}$ is irreducible, then $p$ is also prime.    Irreducible is defined as follows: $n \in \mathbb{Z} $ is irreducible iff the only way to write $n=ab$ for some integers $a$ and $b$ is if $a = \pm1$ or $b = \pm1$. Prime is defined as follows: an integer $n$ is prime iff whenever $a,b \in \mathbb{Z} $ and $p\mid ab$ then it must be true that $p\mid a$ or $p\mid b$ (or both).    Hint: Prove that if $n\mid ab$ but $n$ does not divide $a$, then $\gcd(a,n) = \pm1$. Then use Bézout's theorem and prove $n\mid b$. Search window not opening in ArcMap? Input string was not in a correct format Is "princessship" a real word? Are there any other words which have the same letter 3 times consecutively?