I exported a map and everything looks fine when viewing the PDF. But when printing it on my plotter, all the text disappears. The scale bar would bring but the text associated to the scale bar wouldn't. The title block also disappears. I've tried messing around with the settings. I've checked/unchecked &quot;Embed All Document Fonts&quot; and &quot;Convert Marker Symbols to Polygons&quot;. I'm not sure what I'm doing wrong here. What's weird is I exported this map a few weeks ago and it printed just fine. Today, I needed to make a change to the map and exported it out, and now I am experiencing this issue. I must have changed one of the export settings in the past few weeks but I'm not sure what it is. Has anyone experienced this? Using ArcMap 10.6
I exported a map and everything looks fine when viewing the PDF. But when printing it on my plotter, all the text disappears. The scale bar would bring but the text associated to the scale bar wouldn't. The title block also disappears. I've tried messing around with the settings. I've checked/unchecked &quot;Embed All Document Fonts&quot; and &quot;Convert Marker Symbols to Polygons&quot;. I'm not sure what I'm doing wrong here. What's weird is I exported this map a few weeks ago and it printed just fine. Today, I needed to make a change to the map and exported it out, and now I am experiencing this issue. I must have changed one of the export settings in the past few weeks but I'm not sure what it is. Has anyone experienced this?
Number of finite simple groups of given order is at most $2$ - is a classification-free proof possible? After finishing The Cayo Perico Heist, how long will it be set in hard mode? After I finished The Cayo Perico Heist, I tried setting up another heist in the Kosatka, and got this message :  The setup cost for The Cayo Perico Heist is $25000. Setting up now will set the heist to Hard Mode. Do you wish to proceed?  It seems that there is a cooldown period before the heist reverts to normal mode from hard mode. How long will it take before it reverts to normal mode? Also, does the cooldown timer run even if you're logged out of GTA Online? E.g., if I quit GTA Online just after finishing the heist, the next time I log back in, will I have to wait for the cooldown period to finish before it reverts to normal mode? Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out  the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to  $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&amp;=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &amp;=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$  and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.  Once the solution  $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e.    $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$  However, without knowing this a priori, how would we have approached this problem? Is a hexagonal house feasible? Recently I have been considering the economics and feasability of building a house versus buying a used one. In my design I settled on a hexagon as the logical shape for the frame, which I am considering for masonry construction. This would be a big house, each side being at least 30 feet wide (footprint 2340 sq ft). The advantages of a hexagon are obvious: simple roof shape, no valleys, space efficient, heat efficient (this would be in New England and subject to cold winters), good wind resistance.  What I can't understand is why there seems to be no-one building hexagonal houses? The only notable building I could find at all was the Supreme Court.  Possible explanations would be that the shape is expensive for small houses due to the 6 corners, but for a large house like mine the room shapes work out fine, in fact they are better than a box house. Maybe its just tradition? People don't like it because it looks like church? Builders have some objection to 60-degree angles?  What is the reason there are no hexagonal houses? Gedit has a transparent background since upgrade to 16.10 Since I upgraded to from 16.04 to 16.10, gedit has strange new "feature": its background is totally transparent.    Where can this come from ?  Notice I'm using Unity and don't have any specific gnome package installed besides what comes by default. Notice also that in 16.04 I used the Vivacious theme from ravefinity PPA, but I had to go back to Ambiance as Vivacious was broken in 16.10. Maybe it's related.  [EDIT] changing the 'Fonts and Colors' in gedit to any theme other than default (Kate, Solarized, Oblivion, etc..) solves the issue. So there must be a faulty gtksourview file in my home somewhare, but I can't find it... What does "it" refer to in "it's raining"? Skyrim - Companions attacked me for no reason? Jeffrey Wooldridge in his Econometric Analysis of Cross Section and Panel Data (page 357) says that the empirical Hessian "is not guaranteed to be positive definite, or even positive semidefinite, for the particular sample we are working with.".  This seems wrong to me as (numerical problems apart) the Hessian must be positive semidefinite as a result of the definition of the M-estimator as the value of the parameter which minimizes the objective function for the given sample and the well-known fact that at a (local) minimum the Hessian is positive semidefinite.  Is my argument right?  [EDIT: The statement has been removed in the 2nd ed. of the book. See comment.]  BACKGROUND Suppose that $\widehat \theta_N$ is an estimator obtained by minimizing $${1 \over N}\sum_{i=1}^N q(w_i,\theta),$$ where $w_i$ denotes the $i$-th observation.  Let's denote the Hessian of $q$ by $H$, $$H(q,\theta)_{ij}=\frac{\partial^2 q}{\partial \theta_i \partial \theta_j}$$  The asymptotic covariance of $\widehat \theta_n$ involves $E[H(q,\theta_0)]$ where $\theta_0$ is the true parameter value. One way to estimate it is to use the empirical Hesssian  $$\widehat H=\frac{1}{N}\sum_{i=1}^N H(w_i,\widehat \theta_n)$$  It is the definiteness of $\widehat H$ which is in question. On Ubuntu 12.04 LTS on a Dell Vostro 2420 laptop, each time I start Ubuntu, the screen brightness is set to the maximum level, which gives a high glare. (I have recently installed Ubuntu on the laptop, but did not change any default settings related to brightness, IIRC.) I am currently lowering the brightness manually each time after starting, using the System Settings icon on the left of the screen. Looking for a permanent way to set it to a lower value. Finding an expression for $\tan (a+ b + y)$ which involves only $\tan a, \tan b,$ and $\tan y$. Term for being unable to see glaring errors after working for some time on a task? Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? What is the grammar behind "Thanks be to God"? Would someone give a thorough explanation of 1.14 Minecraft's /data modify? Uninstall apps locked by phone vendor I'm using a Samsung I9000 and I have some annoying apps that came pre installed by the phone vendor, like a music store, a custom web browser and other memory leaking apps.  Those crappy apps can't be uninstalled neither by Manage applications tool, nor by editing the apps tiles in the front menu.  I guess I have to be a root user or something to get rid of those apps, but I don't know how. .bash_history does not update in Git for Windows (git bash) I am talking about when you click on the icons in the dock This command gsettings set org.compiz.unityshell:/org/compiz/profiles/unity/plugins/unityshell/ launcher-minimize-window true  which works for Unity Launcher, (obviously) does not work.  Related to: What is a predicate in c#? Let $G$ be a group, $H\subseteq G$ a subgroup and $a\in G$ an element of the group. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$?   If $H$ has finite index or finite order, this is not possible. I do not understand how dummy variables work in math.   Suppose we have:  $$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$  How is this equivalent to:  $$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$  How does this dummy variable system work?   Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$  ??  Thanks!