I need to know when the user clicks on the button that triggers window.open if there is stable API/way to know beforehand if the user actively has a pop-up blocker?  In some cases the user doesn't know/pay attention that they have pop-up blocker (that block the new window). I would like to inform them by some dialog/or something to authorize a new window by clicking on allow.
Occasionally, I've come across a webpage that tries to pop open a new window (for user input, or something important), but the popup blocker prevents this from happening.  What methods can the calling window use to make sure the new window launched properly?
"You must wear a suit TO an interview" vs "You must wear a suit FOR an interview" In the sentence-     You must wear a suit to an interview   shouldn't the to be replaced by for? Or what's the difference between these two here? The USB has various purposes for data transfer, movement of input and output signals for hardware like connecting an Android device to another Display screen (USB on the go), charging the device etc.(during a device charge the voltage drawn from the USB is same irrespective of the device or OEM then Why are two USB Cables different ?  A USB of one manufacturer may not be supporting another manufacturer. We get problems if we do connect them like      or $n$-th digit in the sequence of natural numbers How can I invest in country A to take advantage of their great interest rate while I live in country B? If I live in the US, for example, where current interest rates are practically nil, how can I take advantage of the fact that ? How to avoid getting shocked by static electricity? sometimes I get "charged" and the next thing I touch something that conducts electricity such as a person, a car, a motal door, etc I get shocked by .  I'm trying to avoid this so if I suspect being "charged" I try to touch something that does not conduct electricity (such as a wooden table) as soon as possible, in the belief that this will "uncharge me".   Is it true that touching wood will uncharge you? How and when do I get charged? I noticed that it happens only in parts of the years, and after I get out of the car... Mean absolute error OR root mean squared error? Why use Root Mean Squared Error (RMSE) instead of Mean Absolute Error (MAE)??    Hi  I've been investigating the error generated in a calculation - I initially calculated the error as a Root Mean Normalised Squared Error.  Looking a little closer, I see the effects of squaring the error gives more weight to larger errors than smaller ones, skewing the error estimate towards the odd outlier. This is quite obvious in retrospect.   So my question - in what instance would the Root Mean Squared Error be a more appropriate measure of error than the Mean Absolute Error? The latter seems more appropriate to me or am I missing something?  To illustrate this I have attached an example below:                    The scatter plot shows two variables with a good correlation,  the two histograms to the right chart the error between Y(observed ) and Y(predicted) using normalised RMSE (top) and MAE (bottom).     There are no significant outliers in this data and MAE gives a lower error than RMSE. Is there any rational, other than MAE being preferable, for using one measure of error over the other? How map multiple inputText to an array property? Twin Paradox, calculating spacetime intervals from both perspectives I've very recently started to try to understand special relativity. I'm want to get a decent understanding of the twin paradox. I'll post what I've done so far and highlight what's gone wrong for me.  The situation is that Alice and Bob both begin at point $x_1$ in Alice's coordinate system $(x,t)$ (we have orientated the axis so that y and z will not matter). Bob then instantaneously moves off with velocity $v$ in the positive $x$ direction. Bob's coordinate system is now $(x',t')=(\gamma (x-vt), \gamma (t-v\frac{x}{c^2}))$. In Alice's coordinates, Bob reaches point $x_2$ after having moved distance $d$, then instantly turns around and travels back to Alice at $-v$. I want to show that Alice's space time interval $\Delta s_A^2$ is greater than Bob's $\Delta s_B^2$, as each path's space time interval is proportional to the proper time passed along that path.    According to Alice: $\Delta s_A^2=-c^2\Delta t^2 + \Delta x_A^2$ where $\Delta t=\dfrac{2d}{v}$ and $\Delta x_A=0$ as she didn't move, but the time elapsed is the time it took bob to travel the distance and then come back. Thus $\Delta s_A^2=-\dfrac{4c^2d^2}{v^2}$  Still according to Alice, Bob's $\Delta s_B^2=-c^2\Delta t^2+\Delta x_B^2$, where the change in time is the same, but now bob has moved distance $d$ twice, thus $\Delta x_B=2d$ ($\Delta x$ refers to total distance traveled rather than displacement which is 0 in this case). Now we have $\Delta s_B^2=-\dfrac{4c^2d^2}{v^2}+4d^2$.     The |size| of Bob's spacetime interval now definitely smaller than Alice's, and this would be all OK, except that when I do the calculations in Bob's frame, they don't agree. This is contradictory to the fact that $\Delta s^2$ is conserved under Lorentz transformations.    According to Bob: $\Delta {s'}_A^2=-c^2\Delta t'^2 + \Delta {x'}_A^2$ where $\Delta t'=\dfrac{2d'}{v}$. I'm not sure if I'm right in saying that $v$, the relative velocity of the two frames, is the only velocity upon which both Alice and Bob will agree on, other than the speed of light. Anyway, $d=\gamma d'$ as lengths contract by $\gamma$ i.e. $d'$ is smaller than $d$ by factor $\gamma$. Also, $\Delta {x'}_A=2d'$, so we have $\Delta {s'}_A^2=-\dfrac{4c^2d'^2}{v^2}+4d'^2=-\dfrac{4c^2d^2}{\gamma ^2 v^2}+\dfrac{4d^2}{\gamma ^2}=4d^2\dfrac{v^2-c^2}{\gamma^2 v^2}\neq \Delta s_A^2$  Although I haven't included them here, my calculations for $\Delta {s'}_B^2$ and $\Delta s_B^2$ agree. Sorry for the ultra-long post, but any help would we well appreciated! Grub rescue prompt, repair grub I won a Samsung Galaxy S6 Edge, signed into it with my Google account, played around with it, wiped it, and sold it.  Now the person can't get past setup because it's saying he must login with an account previously synced to the phone.   Is there anything I can do from my side to remove this lock? I have a firewall with these simple rules:  iptables -A INPUT -p tcp -s 127.0.0.1/32 --dport 6000 -j ACCEPT iptables -A INPUT -p tcp -s 192.168.16.20/32 --dport 6000 -j ACCEPT iptables -A INPUT -p tcp --dport 6000 -j REJECT   Now, suppose I am using TCPDUMP like this:  tcpdump port 6000   And I have host 192.168.16.21 trying to connect to port 6000.  Will/should tcpdump output some packets coming from 192.168.16.21? How do I insert a line at the top of a text file using the command line? Any Set with Associativity, Left Identity, Left Inverse is a Group Related Link:    Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra     I will present my proof (distinct from those in the link) for critique and then ask my question.  $G$ is a set and $ \times $ is an associative binary operation.  Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$.  Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.  $a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$  Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.  As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.  For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.  Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.  In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly.  Specifically, the identity in the second axiom is not well-defined.     Let $(G, *)$ be a semi-group. Suppose    1. $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;    2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group?    This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? –  Derek Holt Sep 17 '11 at 15:31   Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,  when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} =  e_1, aa^{-1}_2 = e_2$?  I think my wording eliminated the ambiguity.  It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes. Mouse Pointer Locator Due to a stroke, I have had some vision impairment in that I sometimes have a hard time locating my mouse pointer on the screen.  I have tried making the mouse pointer larger which helps but is still not the solution.  I am new to the Mac world and would appreciate your help.  In Windows, there is an option where pressing the CTRL key will help to identify the mouse pointer using a temporary display of several radiating concentric circles around the mouse pointer.  Is there something like this on OS X? Does Windows Defender update if Windows Update is set to manual? PSTricks error "Undefined control sequence  \c@lor@to@ps" I have a problem with this file tex:  \begin{document} \begin{figure} \begin{center} \begin{pspicture}(0,0)(1,3.8) \psline{-&lt;}(0,0)(0,.2) \psline{-}(0,.2)(0,.4) \psline{*-&gt;}(0,.4)(0,.9) \psline{-*}(0,.9)(0,1.4) \psline{-&lt;}(0,1.4)(0,2.4) \psline{-}(0,2.4)(0,3.4) \psline{*-}(0,3.4)(0,3.6) \psline{&gt;-}(0,3.6)(0,3.8) \rput(1,.4){$y=0$}%testo \rput(1,1.4){$y=1$} \rput(1,3.4){$y=3$} \end{pspicture} \end{center} \caption{} \end{figure} \end{document}   but I don't understand where is the error. Is there any continuity between James Bond movies that have different actors portraying the title character? There was a foolish (in my opinion) question asked on Scifi.SE       James bond keeps returning with a different face, albeit usually a similar one, and often operates with high-tech gadgetry that seem to be on par with the Doctor's sonic screw driver.   In my opinion each new actor represents a reboot of the franchise using the new face, not a continuation of the story.   Is there any continuity between versions of James Bond which change actors? I'm not talking about characters like M or Q, I'm talking about events or past advisories being referenced in a different Bond movie where the main actor was someone else. How do you link to applications in the Android market? When posting app links, I'm always using a link on Cyrket or AppBrain -- becuase I can't get a link to anything in the official Android Market.  But I have seen some links out there for use w/ Android itself ( ''market://'' protocol ):  market://details?id=de.schildbach.oeffi   (as seen @ )  Is it possible to web-link to apps in the Market so that a non-Android user can link to the appropriate Market page in a non-Android web browser? Which of these two is more correct and why?   The sun shines on my little tree. The sun shines upon my little tree.   From what I've read in dictionaries both are acceptable, however I would like to know if there are subtleties that might arise due to, for example, the context. Prime factors + number of Divisors I know that one way to find the number of divisors is to find the prime factors of that number and add one to all of the powers and then multiply them together so for example  $$555 = 3^1 \cdot 5^1 \cdot 37^1$$  therefore the number of divisors = $(1+1)(1+1)(1+1) = 2 \cdot 2 \cdot 2 = 8$  What I do not know (and can't seem to find when I look) is WHY this relationship exists or a formula which shows its proof.  Has anyone seen such a formula?