I have a win10 and ubuntu 16.04 dual boot setup.  However I mostly use win10 and only use ubuntu for certain tasks.  Is there anyway to suppress the ubuntu bootloader when starting up and only bring it up through activating windows boot options (like pressing F12)?  This would be very helpful as it would save time the 9 times out of 10 that I just continue into windows.
I have a Win 7/11.10 dual boot system with Grub 2 as the default bootloader. However, I want to make Windows bootloader the default bootloader (grub 'dos' like interface scrares the hell out of my sister) so when I start the PC I will have the following boot options: 1. Windows 7 2. Ubuntu  NB: A process to reverse this operation so Grub can be the default bootloader again will also be appreciated
This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this? I am running an analysis on the probability of loan default using logistic regression and random forests.    When I use logistic regression, the prediction is always all '1' (which means good loan).  I have never seen this before, and do not know where to start in terms of trying to sort out the issue.  There are 22 columns with 600K rows.  When I decrease the # of columns I get the same result with logistic regression.   Why could the logistic regression be so wrong?  **Actual from the data**  0 :   41932  1 :   573426  **Logistic regression output**   prediction for 1 when actually 0: 41932 prediction for 1 when actually 1:573426  A**s you can see, it always predicts a 1**   **Random forests does better:**  actual 0, pred 0 : 38800  actual 1, pred 0 : 27  actual 0, pred 1 : 3132 actual 1, pred 1 : 573399 I have been thinking about ways of teaching electronics and I'm wondering if the following is true...  For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?  So, all other things being equal, will the electrons coming out of a 4 volt battery will have twice the velocity of electrons coming out of a 1 volt battery because of ${1 \over 2}mv^2$?  This would seem to imply that the voltage drop across different parts of the circuit is essentially a comparison of the electron speeds at those points.  Furthermore doesn't this imply that the electrons are nearly halted by the time they reach the positive terminal of the battery? This makes intuitive sense to me, it seems pedagogically sound, and it provides an explanation for how electrons "know" to give up their energy across the circuit -- electrostatic repulsion communicates later resistances to earlier parts of the circuit.  How much of the above is correct? The one things that makes me suspicious is that, in this picture, electron density at the positive terminal of the battery is very high, and repulsive forces would prohibit this. On the other hand, if the Joules per Coulomb are not coming from kinetic energy, then what? The field? But wouldn't that just dissipate as light?  Something is wrong but I don't know what. Is it possible to animate the speed of a strip? I want to accelerate the speed of a video and decelerate some seconds before it ends. I wonder if it's possible to do it into the video editing. I know how to increase the speed of a strip but, what I need is a trasition between normal speed and fast speed. Can I disable charging peripherals via USB? I have an external DAC with it's own source of power. However when I connect it to the phone, it drains the phones battery. Is it possible to disable charging via USB on a rooted phone? Or at least slow it down?  My phone is Samsung Galaxy S3, Android 6, CyanogenMod Mac equivalent of Ctrl + Space to select an item on a list I want to select non-consecutive items on a list, without using a mouse/trackpad - there should be no clicks involved.  In Windows and Linux (at least Ubuntu) an item can be selected using Space/ button, which, in combination with arrows and Ctrl key allows me to select multiple items in any environment (file manager, IDE etc.) - e.g.  Enter a directory -> hold Ctrl -> down arrow twice -> press Space -> down arrow twice -> press Space.  Now I should have items 1, 3 and 5 selected. Is there a way to do the same thing in OS X?  PS. Sorry for such a verbose description, but so far every answer I encountered advices using Command + Click, and this is something I don't want to do. How rare are magic items and how many should I be handing out? Benefits and Drawbacks of Swapping Associations of Positivity and Negativity of the Proton and Electron For historical reasons, I understand that a proton's charge was assigned the label of "positive", the electron's charge was assigned the label of negative, and, crucially, equations of current in E &amp; M are written with the notion that current is the flow of positive charge despite the rarity of actual flows of protons in real systems (which I understand are mostly biological systems).   If the possibility existed to instantaneously and at no cost, retcon these associations such that protons have a negative charge, electrons have a positive charge, and current remains the flow of positive charge, would there be any beneficial or detrimental outcomes?  For instance, would there be:   new interpretations of physical laws that would provide different or better insights into any relevant subfield of Physics significant simplifications to notation in any field of Physics or Engineering significant increases in understanding in the fields of electricity, electronics, electrical engineering, radiation/radiative transport, etc. a noticeably easier learning curve for any subfield of physics dealing with the concept of charge   Additionally, is there a permutation of positive charge, negative charge, and flow of current definition that would improve upon the one I suggested? The infinite product of the sines of all positive integers is zero I tried to prove that $\prod\limits_{k = 1}^\infty  {\sin k = 0}$  Define $$P(m)=\prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  then I take the logarithm of both sides $$\log P(m)=\log \prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  which can be written as $$a_m=\log P(m)=\sum\limits_{k = 1}^m  {\log\,\left|\sin k\right|}$$ for any $k\in\mathbb{N}$ we have $0&lt;\left|\sin k\right|&lt;1$, so all terms of the sum above is strictly negative.  I say that the sequence diverges: $\left\{a_m\right\}_{m\in\mathbb{N}}\to\,-\infty$  This is a delicate point on which I have some doubts.  I know that  $$\int\limits_0^\pi  {\log \left|\sin x\right|{\text{ d}}x}  =  - \pi \log 2$$ and that the function $\log \left|\sin x\right|$ is periodic with period $\pi$,  so the integral from 1 to $\infty$ diverges to $- \infty$. Then the series diverges as well for the integral test for convergence.  As $\log P(m)\to\, -\infty$ we can say that $P(m)\to\,0$ as $m\to\,\infty$  Therefore $$\mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\left| {\sin k} \right| = \mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\sin k = \prod\limits_{k = 1}^\infty  {\sin k = } 0} }.\quad \blacksquare$$ Why gpedit and the corresponding registry entries are not synchronized? Let us look at some sample data for 5 hypothetical subjects.  library(emmeans) library(lme4)  # generate some sample data # condition (Placebo, Treatment) # type (some factor, e.g. two different medications) # value (outcome of interest) # subject  data &lt;- data.frame(condition = as.factor(rep(c(0,1), times = 10)), value = rnorm(20), type = rep(c("A","B"), each = 10), subject = as.factor(rep(rep(1:5, each = 2), times = 2)))   Now we like to conduct a mixed model and post-hoc tests with emmeans, where we test conditions per type in a pairwise fashion:  # caluculate a mixed model fit_data &lt;- lmer(value ~ condition*type + (1|subject), data = data)  # emmeans post-hoc test emmeans(fit_data, pairwise ~ condition | type)   So for each type (A or B), I have paired data (condition 0 has 5 data points and condition 1 has 5 data points). If I would do a paired t-test like so:  # paired t-test t.test(data[data$condition==0 &amp; data$type=="A","value"], data[data$condition==1 &amp; data$type=="A","value"], paired = TRUE) t.test(data[data$condition==0 &amp; data$type=="B","value"], data[data$condition==1 &amp; data$type=="B","value"], paired = TRUE)   I would get degrees of freedom of 4 for the paired t-test, but emmeans says the degrees of freedom are 12. Why is there this huge difference? If the emmeans package also would use df = 4, then the p-values would also be more comparable. how can i show site logo, name and slogan Separately in custom area in my site theme, I now use this code to display theme, buts this code don't allow me to edit each of them Separately.  {{ page.header }} Why did the snake wink at Harry at the zoo? This is mostly a memory question - I'm having a hard time remembering this and I can't read the book right now.  On the first book, a snake at the zoo winks at Harry. For the sake of the question, we'll assume that snakes can in fact wink.  I have a fuzzy memory here - I think that it winked before Harry had any kind of interaction with the snake (was this before or after Harry broke the glass?)  So, why did the snake wink at Harry? Given that they had not met ever before.  I suppose that the only reason it would wink at Harry would be because it knows Harry can communicate with it. But this would imply that snakes are capable of sensing such ability on wizards - is there (other) evidence that this is true? Recalculating full extent of File Geodatabase feature class? Having changed the scene units and unit scale, how can I change the default size of new objects? I am using blender most of the time to model things for my 3d printer. So a default unit in "meters" is not really helpful!  Within the scene settings it is simply possible to set the units to millimeter and the unit scael to 0.001 so everything seems to work as I want it to work...   But if I add a new primitive it uses its old default sizes in meters... so a new cube get's 2000 x 2000 millimeters instead of 2x2 or whatever will be useful..  Same for all other new objects...  How can I change the default sizes of new objects permanently?   And how can I change the settings in the application templates? Right now using ctrl+u changes the "general" template - perhaps because I loaded it before...   Thanks a lot and best regards Stefan How can I determine if a used game is pirated? Frequently when I ask a question I have to include quite a bit of code for reference purposes.    It would be nice to be able to bold the important lines in the code.  For example this code:  &lt;DataTrigger Value="True"&gt;     &lt;DataTrigger.Binding&gt;         &lt;MultiBinding Converter="{StaticResource DisableWorkItemConverter}"&gt;             &lt;Binding ElementName="MainForm" Path="PickedWorkItemID"/&gt;             &lt;Binding Path="Id"/&gt;         &lt;/MultiBinding&gt;     &lt;/DataTrigger.Binding&gt;     &lt;Setter Property="IsEnabled" Value="False"/&gt;     &lt;Setter Property="IsSelected" Value="False"/&gt; &lt;/DataTrigger&gt;   Is not way long, but the important part (as seen in ) is the second to last line.  However, if I try to bold it, it ends up looking like this:  &lt;DataTrigger Value="True"&gt;     &lt;DataTrigger.Binding&gt;         &lt;MultiBinding Converter="{StaticResource DisableWorkItemConverter}"&gt;             &lt;Binding ElementName="MainForm" Path="PickedWorkItemID"/&gt;             &lt;Binding Path="Id"/&gt;         &lt;/MultiBinding&gt;     &lt;/DataTrigger.Binding&gt;     &lt;Setter Property="IsEnabled" Value="False"/&gt;     **&lt;Setter Property="IsSelected" Value="False"/&gt;** &lt;/DataTrigger&gt;   Not what I was going for.  It would be nice to be able to draw attention to specific lines of code when needed.    (One suggestion of a different color background sounds nice too (ie a yellow background to draw the eye to the relevant code.)  If there is some other way to bold in code please let me know. How far can one get in analysis without leaving $\mathbb{Q}$? Suppose you're trying to teach analysis to a stubborn algebraist who refuses to acknowledge the existence of any characteristic $0$ field other than $\mathbb{Q}$.  How ugly are things going to get for him?  The algebraist argues that the real numbers are a silly construction because any real number can be approximated to arbitrarily high precision by the rational numbers - i.e., given any real number $r$ and any $\epsilon&gt;0$, the set $\left\{x\in\mathbb{Q}:\left|x-r\right|&lt;\epsilon\right\}$ is nonempty, thus sating the mad gods of algebra.  As @J.M. and @75064 pointed out to me in chat, we do start having some topology problems, for example that $f(x)=x^2$ and $g(x)=2$ are nonintersecting functions in $\mathbb{Q}$.  They do, however, come arbitrarily close to intersecting, i.e. given any $\epsilon&gt;0$ there exist rational solutions to $\left|2-x^2\right|&lt;\epsilon$.  The algebraist doesn't find this totally unsatisfying.     Where is this guy really going to start running into trouble?  Are there definitions in analysis which simply can't be reasonably formulated without leaving the rational numbers?  Which concepts would be particularly difficult to understand without the rest of the reals? This seems to be a chicken-egg problem. The most common task using sudo is installing and removing software.  sudo apt-get purge &lt;appname&gt;   But sudo itself can be removed.  sudo apt-get purge sudo # Do not run this command on production computers!   This is where the fun comes  ubuntu@ubuntu:~$ sudo bash: /usr/bin/sudo: No such file or directory   Although it's obvious that no person in his right mind will purge sudo (other than me), someone can be fooled to run this command (not directly, in its hex mode, or whatever it's called) or a person could SSH in disguised as tech guru and do the mess.  So is there a way of reinstalling sudo? Intuitive proof of $\sum_{k=1}^{n} \binom{n}{k} k^{k-1} (n-k)^{n-k} = n^n$ Is there an intuitive way, though I am not sure how to find a conceptual proof either, to establish the following identity: $$\sum_{k=1}^{n} \binom{n}{k} k^{k-1} (n-k)^{n-k} = n^n$$ for all natural numbers.  I am thinking about binomial formula $$\sum_{k=0}^n\binom nk x^{n-k}y^k=(x+y)^n$$ but I'm not sure how to use it.  I find this problem tantalizing because it looks as if there should be some sort of intuitive way so that is why I share it here. I am looking for an answer like  if possible. Can you find one?