So I have an LGK4 android phone with 8GB of storage, and lately I've been having really weird storage problems; When I try to install or update an app I can't because the storage is full; thing is, I've already cleared my cache, deleted unnecessary apps, and all my photos/videos etc are on the SD card. The apps I do have installed all come to 3.16GB, which theoretically should be the only things on the internal storage, but it still says I just have 611MB available? What happened to those 4GB? To top it all off, I can't access any files on the internal storage, it literally doesn't have any icon or button to access it, so I can't see what's causing this. Any ideas? Should I do a factory reset?
I have critical low storage level and can't understand what takes much of it and how that's possible that the installed applications can use 7 GB? I only use a few social media clients without any games or soft like Photoshop. Also on the second screenshot it indicates that the apps are using just 1.13 GB and so I can't understand the situation. Also what is the "Other"?
What is the difference between parseInt() and Number()? Is it possible to edit the Audit fields on UPDATE? I am trying to customize the legend for this graph. First of all, I would like the legend width to be equal to the x-axis width. At the moment, I am setting the width of the legend via      /tikz/every even column/.append style={column sep=1.0cm}   and I could manually adjust this value so that length equals the x-axis width, but I guess that there is a more elegant way to set the width of the legend.  Second, I don't really like the little color rectangles in the legend. I only want one rectangle in front of every entry and not two. This rectangle does not need to have a border and should be vertically centred with the text. The alignment seems to be okay for the rectangle closer to the text, but not for one the far left.   Update: Thanks to Christian Feuersänger's comment, there is possibility to change the rectangles via   legend image code/.code={%     \draw[#1, draw=none] (0cm,-0.1cm) rectangle (0.6cm,0.1cm);   or what Gonzalo Medina proposed  legend image code/.code={%     \draw[#1, draw=none, /tikz/.cd, bar width=3pt, yshift=-0.2em, bar shift=0pt] %     plot coordinates {(2*\pgfplotbarwidth,0.6em)};   Third, I would like to the legend entries to start next to the text "Legend:".  I don't really know if with these requirements it still makes sense to use legend or if it would be better to add everything manually via after end axis/.code and \ref{label}.  Graph    MWE  \documentclass{article} \usepackage{pgfplots} \pgfplotsset{compat=newest}  \begin{document} \begin{tikzpicture}     \begin{axis}[                 ybar,                 height=6cm,                 title={Some plot},                 width=12cm,                 axis on top,                 y axis line style={opacity=0},                 axis x line*=bottom,                 tick align=inside,                 tick style={draw=none},                 xtick=data,                 ymajorgrids,                 major grid style={draw=white},                 xtick={1,2,3,4},                 legend style={                     at={(0.0,-0.15)},                     anchor=north west,                     legend columns=-1,                     /tikz/every even column/.append style={column sep=1.0cm}                         },             ]                    \addlegendimage{empty legend}             \addlegendentry{\textbf{Legend:}}                                \addplot coordinates {                 (1, 0.2064)                 (2, 0.3961)                 (3, 0.5961)                 (4, 0.7961)}; \addlegendentry[text width=45pt, text depth=]{legend a}             \addplot coordinates {                 (1, 0.3064)                 (2, 0.4961)                 (3, 0.6961)                 (4, 0.8961)}; \addlegendentry[text width=65pt, text depth=]{lengend b}     \end{axis}  \end{tikzpicture} \end{document} I have several AppExchange apps installed that are failing tests. What is the recommended approach to resolving this? Are tests run when a new package is installed? Is there a way to determine what is actually failing besides the name of the method and something like "expected true, actually false"? The test results rarely show enough information to be useful.   What I'm doing atm, is checking for updates for the packages, and then writing the company with the failing test results.   Any feedback would be appreciated. Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ random variables. What is the distribution of $X_1^2 + X_2^2$? Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ which means (mean = 0, variance = $\sigma^2$) random variables. What is the distribution of $X_1^2 + X_2^2$?  My approach is that  $X_1\sim N(0, \sigma^2)$ and $X_2\sim N(0, \sigma^2)$  Then $X_1^2$ and $X_2^2$ have chi-squared distribution with 1 degree of freedom. (I am not sure the degree of freedom and not sure how to show it as well(please help on this))  Then I found the moment-generating function for $X_1^2$ and $X_2^2$;$$m_{X_1^2} = (1-2t)^{-1/2}$$ and $$m_{X_2^2} = (1-2t)^{-1/2}$$  So the moment generating function for $X_1^2 + X_2^2$ is $$m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}$$  So $X_1^2 + X_2^2$ has a chi-squared distribution with 2 degrees of freedom. Is this correct? Time limit on accepting an answer Why is there a 15 minute time limit on accepting an answer in StackOverflow? Resize multiple layers around their respective anchor points I have 50 or so layers (imported images as smart objects) that I would like to scale down, while they stay in place. Is there a way of doing this? I know After Effects has, is there a similar Scale property in photoshop? Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n&gt;2$? How can edge rings be selected in a predictable way? When trying to select multiple edge-rings Shift-Alt-RMB sometimes loops get selected in the one (green) way or the other (red) way.  I couldn't derive a general rule.   How can two or more face-rings following one direction be selected ?  (Easy to reproduce with a default torus mesh.) Server 2012 Core (no GUI): Remote Server manager not working I am trying cmd --> MMC --> Server Manager and then connecting to my Server 2012 no GUI  but i keep getting the following error The WinRM client cannot process the request  I am unable to upload images right now  This is not related to other post "How to Remote Manage Windows Server 2012 in Server Manager?" because title and description of that post is quite different from what i am asking Let $\mathcal{C}$ be a finite binary code such that for any two sequences $\{a_n\},\{b_n\}\in \mathcal{C}$ if $a_j\not=b_j$ for some $j$ then the concatenation binary sequence $a_1a_2\cdots a_n\not=b_1b_2\cdots b_n$, the Kraft-Mcmillan Inequality says that $$\sum_{i}\frac{N_i}{2^{i}}\le 1,$$where $N_i$ is the number of codes in $\mathcal{C}$ with length of $i$.  If we assume the code is prefix free, we will have an elegant probabilistic proof at  (consider the infinite binary sequence generated by i.i.d. binomial random variable with probability $1/2$, define $E_x$ be the event that the sequence has prefix $x\in \mathcal{C}$, then $\sum_i\frac{N_i}{2^i}\le \sum_i P[E_x]=  P[\cup_i E_x]=1$). But for the general case one will not necessarily have the events $E_i$s disjoint, I'm curious if similar argument exists for the general case. (The problem is an excercise in the probabilistic method of alon, I guess some probabilistic argument should exist)  (My attempt is that consider the product topology $T=\prod_i\mathcal C$ whose open sets generated by $E_x$s where $x$ is the concatenation codes of $\mathcal{C}$, we define a natural product measure $\mu$ over $T$ by $\mu(E_x)=2^{-|x|}$. Clearly $\mu(T)\le 1$ (as $T$ is a sub-topology of $\prod_i\{0,1\}$), I want to show $\mu(E_x\cap E_y)=0$ for $x\not=y\in \mathcal{C}$, if not we can find $E_z$ such that $E_z\subset E_x\cap E_y$ by compactness, if we can show $z$ can be write as concatenation code start with both $x,y$ we will done, but I get stuck at here.) How to decide if an effect is magical for Magic Resistance? Some creatures have Magic Resistance:     The [...] has advantage on saving throws against spells and other magical effects.   We know from  that this works against Turn Undead, but does it work against Stunning Strike, Death Strike, Fey Presence, etc? How do we decide? "Where somebody/something at" ? I'm not a native English speaker, so I would like to have an explanation of that expression: "Where something/somebody at?"  Examples :      Where Brooklyn at ? (interlude in a Lonely Island album)      Where you at ? (seen on a game server, instead of Where are you, I suppose)   Are not they supposed to say "Where is/are ..."? How do you keep parents of floated elements from collapsing? Sylow Theorem Explanation The Sylow theorems say that all $p$-Sylow subgroups are conjugate.   Is that the only thing they are conjugate to? In other words could a $p$-Sylow subgroup also be conjugate to something that is not a $p$-Sylow group?   Or does the fact that every element with order dividing $p$ is in a $p$-Sylow group mean that that is never possible? An incorrect answer for an integral As the authors pointed out in  (p. 2), the following evaluation which was in Gradshteyn and Ryzhik (sixth edition) is incorrect (and has been removed). $$ ''\int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)^{3/2}} \frac{1}{\sqrt{1+  \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}+\sqrt{1+  \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}}}} \mathrm{d}x = \frac{\pi}{2\sqrt{6}}'' \qquad (*) $$ A numerical evaluation gives $0.6663771 \cdots$ on the left hand side and $0.64127491 \cdots $ on the right hand side.  I have not succeeded to correct $(*)$.  Do you have any idea on how to evaluate the above integral? Is there a "sentence stretch" that acts horizontally in a similar way as \setstretch acts vertically? Setting custom UITableViewCells height Let $f$ be a real valued, continuous and differentiable function on the real numbers with $f'$ uniformly continuous. Show that $g_n(x)$ with $$g_n(x)=nf(x+1/n)-nf(x)$$ is uniformly convergent to the derivative of $f$.   I cannot figure out how to solve this problem. If anyone could help me for this problem I would be very grateful. Counting overlapping polygons in GeoJSON to result in new set of polygons?