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What does [param: NotNull] mean in C#?
I have added the png transparent image with the setting and its not showing transparent black background is seen. I have the base color aluminum and i want to add that logo of png on to it. The image is black letter with transparent background why its showing in white letters and black background. May i know where im going wrong any suggestion or help. Expected value of sums Suppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get an ace? Hi all im a noob on Ubuntu/Linux World,  So i need some that give a full tutorial to How to install  from apachefriends.org correctly on Ubuntu 11.10 + all beyond Ubuntu for 32 bit and also 64 bit Desktop Version.. and also i need this:   Make a Folder on Your Home Folder and Linked to htdocs from Apache Web Server Make a GUI interface for Xmapp Start Xampp when Ubuntu Boot Up or Start Ubuntu   (Note: i ask it for both X86 and X64 so people in the further can see the tutorials)    New Data:  What i did was i download from  for now and i also what to say that I'm currently running Ubuntu 11.10 X64 bit for Dell Studio 1558 processor intel i3  i will let you know guys which solution was the best, and it is for a Desktop Version of Ubuntu 11.10 (Laptop 15 inch screen ^_^ ) not Server Ubuntu Prove that if a function's average is nondecreasing then the function is nondecreasing. Let $\omega:\mathbb{R}_+\to\mathbb{R}_+$ such that $\omega(0) = 0$ and $\xi(s) = \frac{1}{s}\int\limits_{0}^s \omega(t)dt$ is nondecreasing, i.e., $\forall s,s': s\leq s'$, $\xi(s)\leq \xi(s')$. Show that $\omega(t)$ is nondecreasing a.e., i.e. find a function $w(t)$ such that $w$ is nondecreasing and $\omega = w$ a.e.  I was able to show the reverse implication easily, that if $\omega$ is nondecreasing then $\xi$ is nondecreasing. This way seems more difficult. Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that $$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$  I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox. If the series continues infinitely then each term is just going to get smaller and smaller.  Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$? Converting FLV to MP4 Using FFMPEG and Preserving the Quality Im trying to use ffmpeg to convert my flv files to mp4 to play them on iOS devices but the converted video has a much worse quality than the original one.  Here is the command i use:  ffmpeg -i input.flv -ar 22050 output.mp4   I would really appreciate if someone could provide me with the best settings for flv to mp4 conversion. Mid 2013 MacBook Pro displaying multi color My MacBook Pro was displaying normally until recently. Now when I boot it up it  displays different colors, sometimes white on the screen, but when I hook it up to a TV screen it displays normally.  I need help as to what could be causing this problem and how to fix it?  So far I've tried resetting NVRAM and reseating the RAM, but this didn't help. I also haven't upgraded any software nor added any new items of hardware, so this change in behaviour cannot be attributed to any known actions on my behalf. Is it possible to level grind in Normal Mode? So in Awakening, we can use Reeking Boxes... but what of Three Houses? I'm not far off from the beginning of the game- I have just recently finished the mock battle, actually, but I noticed this calendar mechanic in the game (and in extension, the activity points stuffs). Apparently, the days just... jump.  With this info, I assume there's only a limited number of days before reaching the so-called War Phase... so this brings me back to the question: are there maps or battles that I can replay again and again so that I can level grind the characters, or are there only a limited amount of maps/battles that I can participate in? latexmk: jobname option and clean up incompatibility? I'm using  in a Makefile to compile my document. Since I want my output filename to be different from the input filename, I use latexmk's option jobname.  The problem is: it seems to be incompatible with latexmk -c to clean temporary files.  Here's the specific entry in my Makefile:  these.pdf: header.tex $(OBJS)     latexmk                 \         -pdf                \         -pdflatex="$(TEX)"  \         -jobname="these"    \         -use-make $&lt;   Input files are header.tex containing the latex instruction header and other latex content files. The output files are these.{aux,bbl,bcf,...}. Then latexmk -c does not remove temporary files: .{aux,bbl,bcf,...} files. Sometimes when adding an object to the scene I see that all options (on toolbar or in window after pressing F6) are disabled. Can anyone tell me when can this happen? What is causing it and how to revert it - make it enabled again. It worked for me when I restarted Blender and reloaded the scene. However I would really like to know what can cause it? Are static class variables possible in Python? My question is, if you create a new domain user account for each of the SQL Server processes, what permissions should be set for each account? Or does the SQL configuration manager actually take care of this, and I just had an unforeseen issue?  I quite often have to setup Microsoft SQL Server and wondered if anyone can provide advice on configuring the accounts the services should run as. IMO this has been vaguely documented by Microsoft, while they point you in the right direction I have never been able to find any concrete examples.  To summarize what I've seen so far:  For simple deployments\development environments it is OK to use the virtual account defaults the installer uses: e.g. NT SERVICE\MSSQLSERVER  Avoid using the SYSTEM account, this is not secure.  For production and in domain environments it's recommended to use either a Managed Service Account, or create a domain user account (not an admin) for each service. Allegedly if you use a domain account at installation time the installer will set any required permissions for you.  If changing the service account on an existing install from a virtual account to a domain account the recommendation is to use the SQL Server configuration manager to set the new service accounts. Allegedly this will set any required permissions for you.  I just tried changing the service account in an existing install to a domain account and it would give me a logon failure until I granted the account log on as service permission, which contradicts the part where the SQL Server configuration manager will set any required permissions. (Although I'm not sure if a GPO may have interfered with setting this local security policy)  Microsoft does provide a list of the permissions that the SQL Server Setup grants .  But it's not clear to me if that is something I should be doing manually for the user I create to run the service as, or whether using the SQL config manager should automatically set these permissions.  SQL Server 2014, Domain Controller is on Windows Server 2008 R2. Moderncv adding date of birth to personal information I'm doing my cv with the moderncv package and I'm using the classic style. I would like to add a date of birth line directly above the adress. Does anyone know how to do this? I've tried various solutions but they all seem to be broken. Only ftps with proftpd on debian The characteristic and minimal polynomial of a companion matrix of a monic polynomial $f \in \mathbb F\left[x\right]$ in $1$ variable $x$ over a field $\mathbb F$ plays an important role in understanding the structure of finite dimensional $\mathbb F[x]$-modules.  It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$.  Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence! I've just got my first follower, and I'm wondering if I can get a second person to follow me. Is the game limited to one or two follower(s) or does it allow for a large number of followers? It really seems handy to me to have some followers protect me when I draw attention from a larger group, or when I try to steal something where I really shouldn't... What is the meaning of the beta-coefficient for an interaction term in a crossover study? I have asked a similar question here:  I am puzzled by the interpretation for an interaction term. In my data my Y is an interval variable with the health outcome of an experiment. I have used an interaction term in which I have interacted the condition with the predisposition of the subject considering health status at base level. They are both categorical variables (factor variables in R).  Now it gets complicated because the Condition was two treatments: in treatment A subjects got the placebo first and then the real medicine whereas in treatment B they go the real medicine first and the placebo second. All it changes is the order.   Health outcome = a + Condition * Health.Base   I have the worst state of health at the base level as my reference category I find that interaction with the Condition is statistically significant but I am not sure how to interpret this.  I use the lm() function of R (although my design looks more like an ANOVA) and in the output I get the b coefficient in an output that looks like this:  ConditionB:Health.Base.So.and.So         (Beta and p-value) ConditionB:Health.Base.Excellent         (Beta and p-value)   A statistically significant interaction term for those in Excellent health at baseline would mean that they are affected by the Condition B more than Condition A compared to the reference category people (Poor health at baseline). Is this right? What does the beta-coefficient represent?   If I would like to examine for each Health category at the base line separately without comparing to a reference category I would have to code each category as a dummy variable. However, in this case I would compare whether membership to a specific health status at the base line significantly changes between conditions compared to those who belong to the other health statuses at the base line. Is this right? Again, what does the beta-coefficient represent?  Would it be right to assume that the choice of the interaction between the Condition and the dummy variables is easier to interpret?  --- EDIT --- R output:  Call: lm(formula = HealthOutcome ~ Condition * HealthStatus,      data = datA)  Residuals:     Min      1Q  Median      3Q     Max  -1.5957 -0.5942 -0.2640  0.4423  2.4423   Coefficients:                                        Estimate Std. Error t value     Pr(&gt;|t|)     (Intercept)                            2.595652   0.053044  48.934  &lt; 2e-16 *** ConditionCondB                        -0.001449   0.077071  -0.019    0.985     HealthStatusSo.and.So                 -0.331693   0.078094  -4.247  2.35e-05 *** HealthStatusExcellent                 -0.836724   0.092692  -9.027  &lt; 2e-16 *** ConditionCondB:HealthStatusSo.and.So   0.137490   0.110612   1.243    0.214     ConditionCondB:HealthStatusExcellent  -0.199787   0.133943  -1.492    0.136     --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 0.8045 on 1059 degrees of freedom   (68 observations deleted due to missingness) Multiple R-squared:   0.16, Adjusted R-squared:  0.156  F-statistic: 40.33 on 5 and 1059 DF,  p-value: &lt; 2.2e-16 I am trying to find a way with QGIS to get the intersection area of more than 2 layers.  All the tools in QGIS take just the 2 input layers.  For example in the image below, I have 4 layers and I want to retrieve just the black area.   Furthermore, I need to retrieve the count attribute and add it to the intersected area attribute table (in this case 4)    I know there is a tool that does the trick in ArcGIS, but is there a QGIS solution? How does the garage system work in GTA V? Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$  I tried by subtracting $I_1$ and $I_2$   $$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so  $$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get  $$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$  So $$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$  Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so  $$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$  Need a hint to proceed further.