I cannot navigate back to question from Image page
Support showing image in original size in overlay
We roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by 6? So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?  Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six? How can I rank observations in-group faster? I have a char in c#:  char foo = '2';   Now I want to get the 2 into an int.  I find that Convert.ToInt32 returns the actual decimal value of the char and not the number 2.  The following will work:  int bar = Convert.ToInt32(new string(foo, 1));   int.parse only works on strings as well.    Is there no native function in C# to go from a char to int without making it a string?  I know this is trivial but it just seems odd that there's nothing native to directly make the conversion. Borel set preserved by continuous map Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a continuous map. Show that if $A$ is a Borel subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is a Borel subset of $\mathbb{R}^m$.   I know that for $A$ open subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is open. Likewise for closed. A Borel subset is a countable unions/intersections of these sets, done repeatedly. From those facts, combined with $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, the statement should be intuitively true.  But to do it rigorously, I want to write $A$ as open and closed sets. But since the countable union/intersection can be done repeatedly, I don't know how I can write it that way. Google page title not matching html  tag Design problem - Implement an approval process on case, which sends email to approver showing some additional info, which should not be visible to case owner/agent (who clicked on "Submit for approval" button).  Possible solution - All that additional info will not be stored on Case (otherwise agent will also be able to see it). I am creating a child object to case, where all that additional info be stored. We can make sure agent does not have access to that child object (lets call it ChildObj).   In the email template used, we can include info about that child object. From what i understand, standard email templates dont allow cross object reference. We probably need to use visualforce email templates.  Questions -  1) Can anyone see/point out any issues with that design? 2) I am very new to salesforce. Can anyone give me any example of how i will access child record's field? I mean there won't be any lookup or anything to child object, right? How do i access a field on the child object? Any clue would be helpful Is it possible to use `find -exec sh -c` safely? I'm trying to use find to echo 0 into some files, but apparently this only works with sh -c:  find /proc/sys/net/ipv6 -name accept_ra -exec sh -c 'echo 0 &gt; {}' \;   But using sh -c with find -exec makes me feel very uneasy because I suspect quoting problems. I fiddled a bit with it and apparently my suspicions were justified:   My test setup:  martin@dogmeat ~ % cd findtest  martin@dogmeat ~/findtest % echo one &gt; file\ with\ spaces martin@dogmeat ~/findtest % echo two &gt; file\ with\ \'single\ quotes\' martin@dogmeat ~/findtest % echo three &gt; file\ with\ \"double\ quotes\" martin@dogmeat ~/findtest % ll insgesamt 12K -rw-rw-r-- 1 martin martin 6 Sep 17 12:01 file with "double quotes" -rw-rw-r-- 1 martin martin 4 Sep 17 12:01 file with 'single quotes' -rw-rw-r-- 1 martin martin 4 Sep 17 12:01 file with spaces  Using find -exec without sh -c seems to work without problems - no quoting necessary here:  martin@dogmeat ~ % find findtest -type f -exec cat {} \; one two three  But when I'm using sh -c {} seems to require some kind of quoting:  martin@dogmeat ~ % LANG=C find findtest -type f -exec sh -c 'cat {}' \; cat: findtest/file: No such file or directory cat: with: No such file or directory cat: spaces: No such file or directory cat: findtest/file: No such file or directory cat: with: No such file or directory cat: single quotes: No such file or directory cat: findtest/file: No such file or directory cat: with: No such file or directory cat: double quotes: No such file or directory  Double quotes work as long as no file name contains double quotes:  martin@dogmeat ~ % LANG=C find findtest -type f -exec sh -c 'cat "{}"' \; one two cat: findtest/file with double: No such file or directory cat: quotes: No such file or directory  Single quotes work as long as no file name contains single quotes:  martin@dogmeat ~ % LANG=C find findtest -type f -exec sh -c "cat '{}'" \; one cat: findtest/file with single: No such file or directory cat: quotes: No such file or directory three    I haven't found a solution that works in all cases. Is there something I'm overlooking, or is using sh -c in find -exec inherently unsafe? Can I partition a HDD for playing Xbox 360 games? I am planning to buy an xbox 360 console. my budget permits me acess to only the 4 GB one which needs extra space for many games' install. I have an external HDD of size 40GB. My question is can I divide it into partitions such thet I can use one for gaming on xbox and the other for data storage. Take $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$ I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.  Proof: Suppse that $|B|&lt;2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $ \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.  What do you think?  Thank you, Shir Knife tool doesn't cut obscured edges I'm using Blender 2.66,  I find that the knife tool (K) doesn't work the way it used to.     Dragging across an obscured edge, even if selected, doesn't cut it. There seems to be no way to activate the option to do a midpoint cut or multiple cuts. Is the op-amp inverting or non-inverting? I am studying the function and characteristic of an op-amp to build a line-following robot with the schematics below. However, in many tutorials, the way to differentiate an inverting and non-inverting op-amp is by looking at the connection to ground via the inverting or non-inverting terminal.  However, in the schematics, the op-amp is considered to have connections to ground, thus in this case, how do I differentiate it then?  I may have understood the whole concept wrongly. Pls help!    EDIT: Please ignore the switch as it may have been drawn wrongly by the supplier. I'm having a bit of trouble proving this with the propositional calculus rules.  $$p \lor (p\implies q)$$ Would someone mind helping me and showing which rules they've used with an explanation! Thanks! $(xy + 1)(yz + 1)(zx + 1) $ is a perfect square if and only if each factor is Do redirection operators always open file descriptors in parallel? 1. Consider snippet#1:  $ cat test.txt &gt; test.txt cat: test.txt: input file is output file   It seems that cat makes its input file descriptor point to test.txt and then when it tries to set its output file descriptor to test.txt it throws the above error. Here it seems that cat is aware of the redirect operator and so proceeds to attempt to set the output file descriptor to test.txt  2. Consider snippet#2:  $ cat 1.txt 1:CAT 2:dog $ sed 's/cat/CAT/g' test.txt 1:CAT 2:dog $ sed 's/cat/CAT/g' test.txt &gt; test.txt $ cat test.txt # Note that test.txt is now empty $   Here we see that sed opens test.txt (last argument) in read mode and at the same time sets test.txt as its file output descriptor. Also the '&gt;' operator overwrites the contents of the file BEFORE sed starts to read from it.   I am aware that commands in a pipeline execute in parallel but have not come across any info on how redirection operators behave. Any supporting links would be helpful. Estimate the population variance from a set of means A different approach in distributing $8$ distinct balls into $6$ distinct boxes Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.  We are well acquainted with the traditional Inclusion-Exclusion principle in solving this and it will come up as $191520$ ways.  I tried in this way:  When we distribute $8$ balls into $6$ boxes we have two possible cases  Case $1.$ There are exactly $2$ boxes with $2$ balls each  Case $2.$ There is exactly $1$ Box with $3$ balls in it.  For Case $1$, the distribution is done by choosing $2$ boxes from $6$ boxes and $2$ balls from $8$ balls and those two balls can be arranged in $2!$ ways. But I require another two balls from remaining $6$ balls to be placed in each of these two boxes which can be done in $\binom{6}{2} \times 2!$ ways. Finally remaining $4$ balls can be distributed in remaining $4$ boxes in $4!$ ways. So total number of ways for Case $1.$ is  $$\binom{6}{2} \times \binom{8}{2} \times 2! \times \binom{6}{2} \times 2! \times 4!=604800$$  For Case $2$, the number of ways is  $$\binom{6}{1} \times \binom{8}{3} \times 5!=40320$$  I am pretty sure the value obtained for Case $2.$ is correct. but I could not find where I went wrong in Case $1.$ induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ I encountered the following induction proof on a practice exam for calculus:  $$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$  I have to prove this statement with induction.  Can anyone please help me with this proof? Java Swing: Displaying images from within a Jar I want to keep Ubuntu but remove Windows. I'm having some trouble proving the following statement using mathematical induction:   $$\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n}      ,\text{ (for all sufficiently large n)}$$   I'm sort of confused because $n^{\frac{3}{2}}$ = $n  n^{\frac{1}{2}}$ which means that there are n of the largest possible root values on the left hand side while the right hand side has values like $\sqrt{1}$, $\sqrt{2}$, etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated.