I am creating a list on SharePoint and want to include a simple field in which users can answer a question by choosing a radio button. When I save my changes I see that Sharepoint has included an extra line (or choice) I simply wanted YES, NO and UNSURE as my options. I do not want an empty field. Any help would be great.
The choice field adds a blank option in all my sharepoint lists. How do I remove this? the column settings does not have a blank option.
Which of those functions are PRFs? Two mice heterozygote and black are reproduced. Find the probability of getting the filium dark and heterozygote Entity Query field condition for Webform submissions 403 error after upgrading to apache2.4 After upgrading from Ubuntu 13.04 server, the www pages won't display themselves.     Error 403 Forbidden You don't have permission to access / on this server.   I've changed access control saves from apache2.2 to apache2.4, but that didn't work. What else could I do? Simultaneous collaborative editing of a LaTeX file Is there any way to favorite/bookmark an answer? Lots of questions are very generic, &quot;share an experience&quot; kinds of questions. In these cases, I am quite often interested in specific answers that may not have gotten that many upvotes. These questions also tend to elicit lots of responses, so going back to find the answers I'm interested in somewhere in the middle of a sea of responses is a pain. Is there a way to favorite or bookmark these answers? Balloons and lifting gases Hydrogen is the lightest element, so it's cable of lifting the most weight in out atmosphere (probably not the best terminology there, but you get the picture)  Would hot hydrogen (in the same sense as hot air) be able to lift even more mass? Would a higher or lower density of hydrogen in a ballon lift more? If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?  Basically what are the physics to balloons and lifting?  (really not sure what to tag this, so if someone else could that'd be great) What's the appropriate reaction to an answer that doesn't attribute the source it's copied from? SQL Server Dependencies Is there an easy way to chase down table/stored procedure/function dependencies in SQL Server 2005+?  I've inherited a giant application with lots of tables and even more stored procedures and functions that are long and interlinked.    At the end of the day is there a way to build a dependency tree?  Ideally what I'm looking for goes in both directions:  For a table/procedure - what depends ON it?:  Show me all the stored procedures that eventually reference it (ideally in a tree view such that sub procedures nest out to the bigger procedures that call them)  For a procedure - what does IT depend on?:  Show me all the procedures and tables that a given procedure will (or could) touch when running.  It seems this tool shouldn't be that hard to make and would be incredibly useful for DB maintenance generally.  Is anyone aware of such a thing?  If this doesn't exist, why the heck not?  The built-in functionality in Management Studio is nice but the information does not appear to be complete at all. This is supposed to be related to the  (my justification for the independence tag). In Williams' Probability with Martingales, 2BCL is proven and then the following is given as an exercise:  Prove $S \doteq \sum_{n=1}^\infty p_n &lt; \infty \implies \prod_{n=1}^\infty (1-p_n) &gt; 0$ assuming $0 \leq p_n &lt; 1$. Hint: Show that $S &lt; 1 \implies \prod_{n=1}^\infty (1-p_n) \geq 1 - S$.   Proving the hint: I tried to find $\{a_n\}$ s.t. $$\prod_{n=1}^\infty (1-p_n) \geq \prod_{n=1}^\infty e^{a_n} = e^{\sum_{n=1}^\infty a_n} \geq 1 - \sum_{n=1}^\infty a_n \geq 1 - S &gt; 0$$ That is, find $\{a_n\}$ s.t.: 1a. $1 - p_n \geq e^{a_n}$ 1b. $a_n \leq p_n$ I thought of $a_n = \ln(1-p_n)$ (a reason why $p_n &lt; 1$, I guess) Is that right? Edit: Actually, assuming my proof is right, is $S &lt; 1$ used in the proof? If so, where?  From Williams book: Sometimes we assume that regressors are fixed, i.e. they are non-stochastic. I think that means all our predictors, parameter estimates etc. are unconditional then, right? Might I even go so far that they are no longer random variables?  If on the other hand we accept that most regressors in economics say are stochastic because no outside force determined them with some experiment in mind. Econometricians then condition on these stochastic regressors.  How is this different from treating them as fixed?  I understand what conditioning is. Mathematically, it means we make all observations and inference conditional on that particular set of regressors and have no ambitions to say that inferences, parameter estimates, variance estimates etc. would have been the same had we seen a different realization of our regressors (such is the crux in time series, where each time series is only ever seen once).  However, to really grasp the difference between fixed regressors vs. conditioning on stochastic regressors, I am wondering if anyone here knows of an example of an estimation or inference procedure that is valid for say fixed regressors but breaks down when they are stochastic (and will be conditioned on).  I am looking forward to seeing those examples! Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits  So following tip from : I multiplied top and bottom by conjugate of numerator  $$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$  $$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$  $$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$  $$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$  $$= \frac{1}{2 \sqrt[3]{x^2}}$$  But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)  UPDATE  I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?) 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