Please help me to view the results of this bmi calculations to two decimal points.  here is my code...  @Override public void onClick(View v) {      double weight;     double height;     double bmi;     String msg = "";       if (field_height.getText().toString().equals("") || field_weight.getText().toString().equals("")){          Toast.makeText(getApplicationContext(), "No Valid Values!", Toast.LENGTH_LONG);      }else {          weight = Double.parseDouble(field_weight.getText().toString());         height = Double.parseDouble(field_height.getText().toString());          bmi = height * height;         bmi = (weight / bmi);     } }
What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.  I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.  I know one method of doing this is to use the String.format method:  String.format("%.5g%n", 0.912385);   returns:  0.91239   which is great, however it always displays numbers with 5 decimal places even if they are not significant:   String.format("%.5g%n", 0.912300);   returns:  0.91230   Another method is to use the DecimalFormatter:  DecimalFormat df = new DecimalFormat("#.#####"); df.format(0.912385);   returns:  0.91238   However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:  0.912385 -&gt; 0.91239 0.912300 -&gt; 0.9123   What is the best way to achieve this in Java?
One of the boys who always give/gives the correct answer is Samuel.   As per rule, one of is followed by plural noun/pronoun and singular helping verb, so we should be using gives in the sentence. But the answer is give. Compute $P(Y<3X)$ using joint PDF I'm given a joint pdf  $f_{X,Y}(x,y)=2e^{-x-y}, 0&lt;x&lt;y, 0&lt;y $  and asked to compute $P(Y&lt;3X)$. To do this, I let $Y=3X$ (the boundary) and found that the region of integration is under this line.  To find $P(Y&lt;3X)$, it seems to me that the integral for this region be written as   $P(Y&lt;3X)=\int_0^\infty \int_0^{3x} 2e^{-x-y} dy dx  $  However this isn't right, as the boundaries are actually   $P(Y&lt;3X)=\int_0^\infty \int_x^{3x} 2e^{-x-y} dy dx  $  Why is this? I'm having a hard time seeing how to construct these boundaries (especially the fact that the $y$ boundary goes from $x$ to $3x$ instead of $0$ to $3x$)  Thank you. In case of shell-builtins (eg type itself):  $ type type type is a shell builtin  $ which type &lt;Doesn't return anything since it's a shell builtin, silently exits&gt;   In case of commands (normally) (eg python):  $ type python python is /usr/bin/python  $ which python /usr/bin/python   In case of which (which is a command located at /usr/bin/which)  $ type which which is hashed (/usr/bin/which) $ which which /usr/bin/which   Why does type which say that which is hashed? What is the significance of which being hashed and what does it actually mean? What is the difference between range and xrange functions in Python 2.X? itoa() is a really handy function to convert a number to a string. Linux does not seem to have itoa(), is there an equivalent function or do I have to use sprintf(str, "%d", num)? How can you change the href for a hyperlink using jQuery? Where are my Screenshots? I pressed F2 to take a few Minecraft screenshots. Later I couldn't find them. Where are they stored on my computer? Using dconf-editor I tried enabling   org -&gt; gnome -&gt; nautilus -&gt; list-view -&gt; use-tree-view     but even after the check box is selected, plus a reboot, it still doesn't switch to the tree view on the side bar.  In previous versions of Nautilus / Ubuntu the side-pane-view setting used to be in the window-state section (which is now missing)    I'm I missing something, or is there no way to enable Tree View on the side bar in 14.04?  Is there a way to manually create the missing entry back in window-state ? I'm a mathematician interested in abstract QFT.  I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator.  I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator.  But I don't see how this forces $T^2=-1$.  (Or maybe it doesn't force it, it merely allows it?)  Here's another version of my question.  There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$.  If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$.  So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$.  (I'm assuming Euclidean signature here.)  My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?  (I posted a  on physics.stackexchange.com last week, but there were no replies.)    EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search.  If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$.  I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator. Maximum number of files being uploaded by input type file Is it (theoretically) physically possible to project an image into thin air? register_globals error in php I was stuck up with the error   directive 'register_globals' is no longer available in PHP in unknown on line 0   when tried to check the php version using "php -v" after enabling register_globals in php.ini file. I am not getting any php version info by doing so. Instead it throws the above mentioned error.After turning off this option, php info works quite well. It is very essential for me to have register_globals to be turned on.How can I have this corrected.  my php.ini is as follows:   ; Default Value: None ; Development Value: "GP" ; Production Value: "GP" ; http://php.net/request-order request_order = "GP"  ; Whether or not to register the EGPCS variables as global variables.  You may ; want to turn this off if you don't want to clutter your scripts' global scope ; with user data. ; You should do your best to write your scripts so that they do not require ; register_globals to be on;  Using form variables as globals can easily lead ; to possible security problems, if the code is not very well thought of. ;  register_globals = On  ; Determines whether the deprecated long $HTTP_*_VARS type predefined variables ; are registered by PHP or not. As they are deprecated, we obviously don't ; recommend you use them. They are on by default for compatibility reasons but ; they are not recommended on production servers. ; Default Value: On ; Development Value: Off ; Production Value: Off ;  register_long_arrays = Off  ; This directive determines whether PHP registers $argv &amp; $argc each time it ; runs. $argv contains an array of all the arguments passed to PHP when a script ; is invoked. $argc contains an integer representing the number of arguments ; that were passed when the script was invoked. These arrays are extremely ; useful when running scripts from the command line. When this directive is ; enabled, registering these variables consumes CPU cycles and memory each time ; a script is executed. For performance reasons, this feature should be disabled ; on production servers. ; Note: This directive is hardcoded to On for the CLI SAPI ; Default Value: On ; Development Value: Off ; Production Value: Off ; register_argc_argv = Off  ; When enabled, the SERVER and ENV variables are created when they're first ; used (Just In Time) instead of when the script starts. If these variables ; are not used within a script, having this directive on will result in a ; performance gain. The PHP directives register_globals, register_long_arrays, ; and register_argc_argv must be disabled for this directive to have any affect. ; auto_globals_jit = On I work with QGIS 2.10.1 on Windows 7 (x64). I have dissolved polygons from an attribute value and the polygons geometries being not perfectly joined together, I get some 'garbage' nodes I need to delete. I have tried to edit the layer and using nodes tool, selecting and deleting the nodes, but it seems impossible to do (nodes are not deleted at all!). Any idea on how to process this kind of clean-up?  EDIT : you can  for testing. The other day I was trying to download something for Minecraft and it involved my bin folder. So I was looking for it and couldn't find it. I then thought that I might have accidentally deleted it, but didn't find it in the trash. I have no idea how to get the bin folder back with all the Minecraft binary files in them. Data   I want operator users on this machine to mount their own cifs shares The sudoers file already contains the /bin/mount -t cifs //*/* /media/* -o username=* command for all operators I want the users to mount a cifs share through a script typing the password only once, not twice. The sudo password and the cifs password are identical.   What I already have  This script works:  #!/bin/bash sudo 'mount -t cifs //192.168.1.1/home /media/$USER/home -o username=$USER'   ...but it requires the users to type the same password twice!     Once for sudo Once for the mount itself   This would also work:  #!/bin/bash echo -n Password:  read -s szPassword echo $szPassword | sudo -S sh -c 'echo $szPassword | mount -t cifs //192.168.1.1/home /media/$USER/home -o username=$USER'   ...but this would require me to allow all operator users to be able to sudo sh (major security problem)  Question  How to mount a cifs share in bash¹ without putting sh in the sudoers file nor creating a permanent/temporary file???  Note 1: no python, perl, C, Go, ... please? Note 2: I know I can just remove the password through the sudoers file, but I'm trying to tighten security, not loosen it, without giving up convenience... Software for measuring internet traffic? I'm interested in finding a free piece of software for Windows XP &amp; 7 that allows us to measure both incoming and outgoing internet traffic, but not traffic between users of the local network.  I live in a shared household with three others, and we're interested to see which of us is using the largest amount of our monthly internet quota. We're all happy to install the necessary software.  Any suggestions? How to link dataframes in ArcGIS or QGIS? The below image shows a layout in ArcGIS 10.1 with four dataframes.  Each dataframe is of the same geographic area, although they are different images. For example, the top left image is a 1989 DOQ, the top right image is a reversed 1989 DOQ, the bottom left image is a 2012 DOQ, and the bottom right image is a reversed 2012 DOQ.   Is it possible to link the dataframes so that panning the image in one dataframe pans the images in all of the other dataframes (i.e. similar to how you can "Link Views" in Erdas Imagine)? Is this possible in QGIS? I have a quick question about updating my references from Bibtex.  I am using TexStudio as my Latex editor.  I have to make changes to some of the references.  I put the changes in the "references.bib" file.  I then compile the Bibtex and the main document I'm working on.  Sometimes the changes show up, and sometimes they don't, and I haven't figured out why. I have even gone as far as trying to delete the references by changing the filename, but the original references still appear.  Here's what I put in my main document:   \newpage %Puts in References  \bibliographystyle{unsrt}  \bibliography{references}   My references.bib file shows all the changes I have made.  Any suggestions are appreciated! Most of the time, I use good and well interchangeably. However, on many occasions I would find well or good a misfit. Please suggest the proper usage. I have meticulously read through similar questions, answers and their comments but I still do not understand how the deletion is done. From what I understand I email the customer service and then they do what exactly?     1) They disable my account so that I can no longer use it?      2) They reassign all of my questions and answers to some other   account? To which one?      3) They delete all my questions and answers so that they go away from   public view?   What is the flexibility (if it exists) of the deletion procedure? Is it possible to erase the traces of my former association with the site. Can I request any of the following be done?     4) All of my questions and answers be deleted from public view?      5) Some of my questions and answers be deleted? If yes what criteria?      6) Can I have all of my contributions be assigned to some other   account which runs as sort of a "collective account" and already has   tons of content so that it doesn't happen like my contributions just   move to some other url but still stay grouped under that location and   therefore that location can be discovered as representing my account?      7) Can I have my contributions dispersed through several "collective   accounts?      8) Can my current public name be removed from any comments to my posts   so that my posts cannot be identified later as once belonging to me?   Please be polite with me. I wouldn't have asked if I have found the answer in the other questions. The other questions are also old and it is not clear what the situation is as per 05/2012.  UPDATE (for those interested):  I asked the Stackoverflow team about the procedure. Apparently none of the points 2-8 I asked above are possible, only 1 will take place. The link to my user page from my contributions is going to be removed and replaced with non-clickable text of an anonymized user name like "user1234". Whether this number is the same as my current user_id is not clear. That's all the information I got.  I'm going now to rename myself to "Anonymous" then request the account deletion. Those interested in the procedure will witness the exact result on this page.  Farewell.