Data splitting in regression
What is the difference between test set and validation set?
Blank screen after installing nvidia restricted driver I've  and rebooted my machine and now I get a black or blank screen.   How can I fix this? The question I'm asking is pretty much:  How do you convert PSI to Velocity?  Everywhere I look it says You can't, but can you?  And what would you need to know to do so if there is no direct way to convert?  My homework is to convert 0.14 PSI into m/s.  EDIT Sorry guys my teacher does like being tricky, no clue why. Sorry for any inconvenience this may have caused. Computing dominant area in polygon using ArcGIS Desktop? "#include" a text file in a C program as a char[] Is there a way to include an entire text file as a string in a C program at compile-time?  something like:   file.txt:  This is a little text file  main.c:  #include &lt;stdio.h&gt; int main(void) {    #blackmagicinclude("file.txt", content)    /*    equiv: char[] content = "This is\na little\ntext file";    */    printf("%s", content); }    obtaining a little program that prints on stdout "This is a little text file"  At the moment I used an hackish python script, but it's butt-ugly and limited to only one variable name, can you tell me another way to do it? How can I upgrade from Mac OS X Tiger 10.4? If I buy a regular retail Snow Leopard disc (i.e. the £25 version), can I use it to upgrade a Mac from Tiger (10.4)?   on the Apple Store’s Q&amp;A site suggests it’ll work, but Apple seem to imply that you need the (significantly more expensive) Mac Box Set (“”). Is there a simple method of parsing XML files in C#? If so, what? Remove the extra headers from the e-mails sent Why does demand for stock rise when a company appears to have high future value? Trouble doing polynomial interpolation The equals() method in Java works unexpectedly on Long data type Let $L,M$ two subspaces of the vector space, $V$ such that both $L,M \ne V$.   Prove: $L\cup M \ne V$.     I think this is a case of a proof by contradiction. Lets assume $L \cup M = V$.   Hence, $$\dim(V) = \dim(L) + \dim(M) - \dim(L\cap M)$$  How to proceed? Minecraft crashes on launch with EXCEPTION_ACCESS_VIOLATION, Problematic frame: ig4dev32.dll or ig4dev64.dll or ig4icd32.dll or ig4icd64.dll When I try to launch Minecraft 1.7.4 or newer, the game crashes on startup, and the game output screen says this:  #  EXCEPTION_ACCESS_VIOLATION (0xc0000005) at pc=*, pid=*, tid=* # # JRE version: Java™ SE Runtime Environment (7.0_51-b13) (build 1.7.0_51-b13) # Java VM: Java HotSpot™ Client VM (* mixed mode, sharing windows-* ) # Problematic frame: # C  [ig4dev32.dll+0x3e88] OR # C  [ig4dev64.dll+0x55ec] OR # C  [ig4icd32.dll+0x104a77] OR  # C  [ig4icd64.dll+0x84930a]   This happens with both Java 7 and 8.  What causes this error, and what can I do to fix the issue? I'm trying to solve this problem from Hatcher 3.3.24.  Let $M$ be a closed connected 3-manifold, and write $H_1(M, \mathbb{Z})$ as $\mathbb{Z^r} \oplus T$ where $T$ is torsion. Show that $H_2(M, \mathbb{Z})$ is $\mathbb{Z^r}$ if $M$ is orientable or $\mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$ if $M$ is not orientable. In particular $r \geq 1$ if $M$ is not orientable. Using exercise 6 construct examples showing there are no other restrictions on the homology groups of closed 3-manifolds. (In the nonorientable case consider the manifold $N$ obtained from $S^2 \times I$ by identifying $S^2 \times \{0\}$ with $S^2 \times \{1\}$ via a reflection of $S^2$.)  I think I've solved the first part, here's my proof:   First note the following theorem: If $M$ is a closed connected and orientable n-manifold then $H_n(M, \mathbb{Z}) = \mathbb{Z}$ and $Tors(H_{n-1}(M, \mathbb{Z})) = 0$. If nonorientable, then $H_n(M, \mathbb{Z}) = 0$, $Tors(H_{n-1}(M, \mathbb{Z})) = \mathbb{Z_2}$.  Now note by Poincare duality and UCT we have that $H_2(M) \cong H^1(M) \cong \mathbb{Z^r} \oplus T_0 \cong \mathbb{Z^r}$.   In the nonorientable case consider the exact sequence:  $0 \longrightarrow H_2(M) \longrightarrow H_1(M) \longrightarrow H_0(M) \longrightarrow 0$  But we know $H_1(M) = \mathbb{Z^r} \oplus Z_2$ from the earlier theorem and $H_0(M) = \mathbb{Z}$.   Therefore $H_2(M) \cong ker(\mathbb{Z^r} \oplus Z_2 \rightarrow \mathbb{Z}) \cong \mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$.   For the rest of the question I'm quite lost. Display Top X results for each unique value I am trying to figure out how I can output any machine that has had the latest 3 backups be larger than 10GB. My SQL Table looks similar to the below table:  ComputerID | BackupImageName | WindowsVolumeGUID | FileSize | TimeCreated 1234       | PCBackup.spi    | kjsdfai-dadad-ddd | 10400    | 05-22-20 1234       | PCBackup1.spi   | kjsdfai-dadad-ddd | 10500    | 05-23-20 1234       | PCBackup2.spi   | kjsdfai-dadad-ddd | 10300    | 05-24-20 1234       | PCBackup3.spi   | kjsdfai-dadad-ddd | 10900    | 05-25-20 1225       | PCBackup4.spi   | kjsdfa5-dadad-ddd | 10600    | 05-22-20 1225       | PCBackup5.spi   | kjsdfa5-dadad-ddd | 5000     | 05-23-20 1225       | PCBackup6.spi   | kjsdfa5-dadad-ddd | 10000    | 05-24-20 1225       | PCBackup7.spi   | kjsdfa5-dadad-ddd | 4588     | 05-25-20   I need to output all machines (computerID's) or Volumes (WindowsVolumeGUID) that have their latest 3 files in the list greater than 10000 file size. So from the above table it would output the computerID of 1234 because its latest 3 are greater than 10000 file size and even though the computerID 1225 has 2 files greater than 10000 it would not output the result because there is a break in-between the two large files. The search will need to take into account the WindowsVolumeGUID since some machines have more than one volume and I need to take into account the individual volume backups and count the latest entries based on that.  Alternatively, I can also accomplish the same thing if there is a way to output the top 3 results for each unique WindowsVolumeGUID and not just across the board.   Any help would be greatly appreciated! Is it possible to colour or put textures on Bevel curves? So with my model, I've been struggling to create hair because it always looked weird or off and recently I discovered the "Bevel Curve" addon and it's an absolute lifesaver.   So I converted the bevels into mesh to colour it, except it lags blender like crazy and I need this model for a video game I'm currently developing called Candles(so I can't upload the model, sorry) but I was wondering, is it possible to put a texture on a bevel curve like you would mesh or any other method or giving it colour? What counts as "observation" in Schrödinger's Cat, and why are superpositions possible? So if I understood correctly, Schrödinger's Cat is a thought experiment that puts a cat inside a box, and there's a mechanism that kills the cat with 50% probability based on a quantum process. The argument is that the cat now must be in a superposition of dead and alive.  This makes sense at first, but the state of the cat inside the box will affect the outside world in an observable way, right? For example if the cat dies, it might meow loudly which would be audible. If it didn't meow, it would produce a thud on the ground when it dies. And even if the ground was very solid, the redistribution of mass inside the box will affects its gravity field which means the whole universe theoretically immediately observe's the cat's death.  So extending this argument to all superpositions, the different states would cause different effects on the rest of the universe, usually a slight change in the gravity field is the minimum. This gravity perturbation would propagate throughout the universe, and even all the experimenters go to sleep with thick, thick earplugs, somebody or something in the universe is going to inadvertently observe the event and the superposition immediately collapses. Thus superpositions cannot exist beyond an extremely short amount of time.  What's wrong with my reasoning? How to recover a deleted file from /data partition? Finding critical points (extrema) directly without using eigenvalues If we have a function $f: \mathbb R^3\rightarrow \mathbb R$ of the form:  $$f(x)=(a_1x_1^2+a_2x_2^2+a_3x_3^2)e^{-x_1^2-x_2^2-x_3^2}$$  where $a_1&gt;a_2&gt;a_3&gt;0$.   We can find the critical points by:  $$\frac{\partial f}{\partial x_1} =2x_1e^{-x_1^2-x_2^2-x_3^2}(a_1-a_1x_1^2-a_2x_2^2-a_3x_3^2)=0$$  And solving symmetrically for $x_2$ and $x_3$. We get critical points: $(0,0,0) , (\pm 1,0,0) , (0,\pm 1, 0) , (0,0,\pm 1)$   We could determine the nature of them by looking at the eigenvalues of the Hessian, but is there a way to just analyse it directly? Obviously $(0,0,0)$ is a strict global minimum, but I can't quite see that the other points are. Any pointers/hints would be appreciated. Are global variables bad? Why can't polynomials have negative exponents or division by a variable Why can't:  $$2x^{-3} - 3x$$  or  $$\frac{1}{2x}$$  be polynomials too?  Why have a definition that excludes these algebraic forms?