How should I partition my drive for my Ubuntu installation? So firstly I'm going to dual boot windows and Ubuntu (Windows already installed) and am wondering how I should partition my other drive, which is completely empty.   Do I just need one massive partition, or smaller partitions?  Its formatted as NTFS or whatever it is, the Windows standard.    I'm planning to install Ubuntu 13.04 and am relatively new, Ive used Wubi but now I want the real experience.   Just one quick bonus question, how can I distinguish between drives in the installation process. I've heard bad stories about choosing the wrong drive. Also I cannot disconnect and hard drives as I'm not going to go fiddle around inside my laptop :)  Thanks in advance :)
What's your recommendation on drive partitioning schemes for a desktop and home server? What partitioning scheme do you recommend for a desktop? I've always created three or four primary partitions -- root, swap, home, and sometimes a separate boot partition. Ubuntu's default install offers LVMs. I've never had to add additional drives or space, so it never seemed like a big deal. Whenever I do a fresh install, though, I always think there might be a better way.
bashrc lazy substitution Why are the endpoint lines of intervals (`|-|`) not shown completely? I use \draw[|-|] () to (); to draw a few intervals.  I expect that each interval is rendered like |----| (see the 2nd and the 3rd intervals in the figure below).  However, the top and the bottom intervals below show only a half of their endpoint lines (i.e., |).  Why are these endpoint lines (|) not shown completely? How to show them?        \documentclass{standalone} \usepackage{tikz}  \begin{document} \begin{tikzpicture}   \draw[|-|, red] (0,0) to (1,0);   \draw[|-|, red, dashed] (0,1) to (1,1);   \draw[|-|, blue, dashed] (0,2) to (1,2);   \draw[|-|, blue] (0,3) to (1,3); \end{tikzpicture} \end{document} Bag A has $3$ white and $2$ black marbles. Bag B has $4$ white and $3$ black marbles.  Suppose we draw a marble at random from Bag A and put it in Bag B. After doing this, we draw a marble at random from Bag B, which turns out to be white. Given this information, what is the probability that the marble we moved from Bag A to Bag B is white?  At first since the probability that white is $1/2$ is $2/5$, and the probability that white is $5/8$ is $3/5$, I thought a sufficient answer would be $$\begin{align}\tfrac 1 2 \cdot \tfrac 2 5 + \tfrac 5 8 \cdot \tfrac 3 5 &amp; = \tfrac 1 5 + \tfrac 3 8 \\ &amp; = \tfrac 8 {40} + \tfrac {15}{40} \\ &amp; = \tfrac {23}{40}\end{align}$$, but that was wrong.   Can someone provide me with an answer? Thanks What is the difference between "I have a doubt with ...." & "I suspect there is a problem"? I tried control+shift+power button as well as command+shift+power button. Neither seem to work. cat files with directory Is there a command to show the directory/file name when cat files?  For example: assume two files f1.txt and f2.txt are under ./tmp  ./tmp/f1.txt ./tmp/f2.txt    Then when I do cat ./tmp/*.txt, only the content of files will be shown. But how to firstly show the file name, then the content?, e.g.:   (The needed command):  ./tmp/f1.txt:    This is from f1.txt  and so on  ./tmp/f1.txt:  This is from f2.txt ...   Is there a command to do it? (There seems to be no option for 'cat' to show the file names) What does the word treats mean in this context? I don't think it means entertainment, food, drink etc.. Maybe it is an old usage of this word? Does it mean influencing factors or something like that?  This is the first part of the title of the first chapter of the novel "Oliver Twist" by Charles Dickens. The whole title reads:     Treats of the Place Where Oliver Twist was Born and of the Circumstances Attending His Birth Application_End global.asax Can anybody tell me when Application_End is triggered in a lifecycle of an application? When all sessions are ended, will Application_End be triggered automatically? + Are there any other reasons why Application_End could be triggered? What would be the best approach to get the recordtype id How bias arises in using maximum likelihood to determine the variance of a Gaussian? While reading , I ran into the following explanation for why there is an error in estimating the variance of Gaussian distribution using Maximum Likelihood. On ,      Illustration of how bias arises in using maximum likelihood to   determine the variance of a Gaussian. The green curve shows the true   Gaussian distribution from which data is generated, and the three red   curves show the Gaussian distributions obtained by fitting to three   data sets, each consisting of two data points shown in blue, using the   maximum likelihood results (1.55) and (1.56). Averaged across the   three data sets, the mean is correct, but the variance is   systematically under-estimated because it is measured relative to the   sample mean and not relative to the true mean.   I'm pasting figure 1.15 and the required equations (1.55, 1.56) here for convenience.  $\mu_{ML} = \frac{1}{N}\sum_{n=1}^{N}x_n$   $\sigma_{ML}^2 = \frac{1}{N}\sum_{n=1}^{N}(x_n - \mu_{ML})^2$    I can imagine how maximum likelihood estimation would look pictorially. However, I'm not able to understand the figure or the explanation. I would appreciate if someone could label the figure or explain what points are considered in each part of the figure and why the curve could look like that in each case.  I looked up , and I'm able to understand why there is a factor of $\frac{N-1}{N}$ between the true $\sigma^2$ and $\sigma_{MLE}^2$.  ${\rm I\!E}[\sigma_{MLE}^2]= (\frac{N-1}{N})\sigma^2 $ Find this integral $$\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$$  My try: let $$f(x)=x^4-2x^3+2x^2-x+1$$ I found  $$f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)$$ so $$I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ so $$2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ then I can't,Thank you very much Why would var be a bad thing? When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$? $$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$ In other words, a rational number is a number that can be written as one integer over another. For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.  Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?  Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.  However, what is this then: $\dfrac {\sqrt 7} {2}$? What is ambient occlusion? I've heard about ambient occlusion and it looks nice, but what exactly is it? Is the phrase "for free" correct? When I place a multipage pdf file into InDesign CC, the pages go outside of my document settings, making this function useless.  What do I have to do to make the placed  pdf pages go within my document settings, which in this case is 11 in. Length X 8.5 in. width, .5 in. margins on all sides, with two columns per page. I would greatly appreciate an answer to this, Thank you. Best Algebraic Geometry text book? (other than Hartshorne) How to show $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ Double Sideband Suppressed Carrier modulator List I am using ring modulator for DSB-SC  am modulation in a transmitter circuit. Can anyone please tell me about more DSB SC modulator for am modulation? I want to know which one to use where? Show that $\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) = \sum_{i=1}^n (x_i - \bar{x})(y_i)$. Show that $\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) = \sum_{i=1}^{n} (x_i - \bar{x})(y_i)$. I fell into this hole where I keep finding that $$\sum_{i=1}^n (x_i - \bar{x}) = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} = n\bar{x} - n\bar{x} = 0$$ This seems true, but I know that I am missing something, since this implies that covariance is always zero, which is obviously not true.