I am new to stock investments. I usually ask for stock broker to write up a quote for me. I can also check the prices on the . However I am not sure which price she quotes,the buy or the sell? My question is what is the difference between the two,If I am selling my shares which price should I look at,and also if I am buying which price do I look at? The questions  I looked at made reference to Bid and Ask which are not terms used on the Malawi Stock Exchange. Can someone clarify this for me please as the terms used in the US may not be the same
In my online brokerage account, I want to buy a particular stock and I see the following:  Bid:  13.20 x200           Ask:  13.27 x1,000   the current stock price is 13.22.   Can someone explain what the bid and ask prices mean relative to the current price?  If I buy 1000 shares, why would I pay more?
I notice that the space between each column of a table in LaTeX is not quite wide (at least for me). Is there any way to set this? I found the \hspace, but I don't get how to use this in the table. Is normal extension of normal extension always normal? I would like to visit Iran as a tourist but I'd also like to visit the USA some time after. I know the relations between the two countries are bad.   Is it likely that I'm not allowed to enter the USA after having been to Iran?   If it matters, I've never been to either of those countries. I'm a Swiss citizen, an atheist from a Christian background. add transistor model to LTspice I'm trying to add the mmbt3904lt1g model to LTspice. - I've copied its data from the MMBT3904LT1.LIB.txt in the standard.bjt library. - I've added an NPN to my schematic then I've changed its value to mmbt3904lt1g  - through the Edit->spice directive I've pasted the path to this library I thought it was all to use this device but no simulation is achieved.. I get only: Can't find definition of model "MMBT3904LT1g"  where I'm doing wrong ? Revision history suggests that I edited the post body, whereas I didn't I was doing some re-tagging and I noticed that the automatically generated revision comment says that I edited the post body where I only edited the tags. Prove that $\exp(x)>0$ using only formal definition of exp Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks How to test @future methods with setup objects and avoid MIXED_DML_OPERATION Is there a way to make Git mark a file as conflicted? I am trying to understand how complex numbers made their way into QM. Can we have a theory of the same physics without complex numbers? If so, is the theory using complex numbers easier? How do I defeat the Kayran in the first chapter of the Witcher 2? In the first chapter you are supposed to slay the Kayran, this enormous beast:    I'm having a hard time with it. I drank the Mongoose potion for poison resistance and I have the Kayran trap, but it just hits me too often with its tentacles. Especially when the Kayran swings a tentacle sideways, not from above, I have a very hard time evading it.  Is there any trick besides not getting hit by the tentacles?  With a lot of luck I'm getting as far as the point where it's throwing stones. Any idea what I'm supposed to do then? I just wanted to make some input where I press a key and the auto-repeat function engages so I don't have to press the 0 key many times to make a long number.  10000000000000000000000000000  This has always worked without fail in the past, but it's not something I do often and I'm not sure where to look next to figure out what might be preventing this.     I've toggled the sliders to change the delay for key repeat, but the only thing I can hang my hat on is that perhaps Lion disables this with the new feature where you hold down some keys (like the vowels, but also some consonants) to get alternate accented characters.  s turns into ß or ś or š and so on....  Am I missing an obvious setting, or is this something that has moved to a hidden setting or is simply a bug that these controls are still around since key repeat is now deprecated?  Comments welcome if anyone doesn't have an answer, but can provide specifics where this does work for them. I've checked two Macs so I'm guessing it's not related to keyboard type or connection method. Moment generating function (MGF) of the ratio distribution $\displaystyle\frac{X}{Y}$ If we know the moment generating functions (MGFs) of the random variables $X$ and $Y$ to be $M_{X}(s)$ and $M_{Y}(s)$, respectively. The MGF of the sum $X+Y$ will $M_{X}(s) \cdot M_{Y}(s)$.  So what is the MGF of the ratio distribution $\displaystyle\frac{X}{Y}$? I read here at  that holding Shift during boot can bring up the GRUB menu. However I try holding one or the other Shift keys, tapping, tapping then holding, nothing works. Am I missing something? How does one access the GRUB menu during boot time?   No dual boot here, just standard Lubuntu 14.04 install. Does the Google spider render JavaScript? How to execute consecutive commands from history? Suppose I want to execute a sequence of four commands that I have executed before. If the first one is 432 in the command-history, then I could do:  $ !432; !433; !434; !435   I'm curious, is there a more efficient way to accomplish this? Why is my network interface named enp0s25 instead of eth0? A lot of my scripts stopped working after installing Ubuntu 15.10 and I have trouble remembering this new name as I was used to ethx and wlanx for years. Is there a reason for this change? Do I have to get used to this or can I just rename it and go back to good old eth0?  $ ifconfig  enp0s25   Link encap:Ethernet  HWaddr 77:5a:5e:a6:86:d5         inet addr:192.168.31.239  Bcast:192.168.31.255  Mask:255.255.255.0       inet6 addr: fe80::725a:b6ff:fea6:86d7/64 Scope:Link       UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1       RX packets:4833 errors:1332 dropped:0 overruns:0 frame:666       TX packets:5589 errors:0 dropped:0 overruns:0 carrier:0       collisions:0 txqueuelen:1000        RX bytes:3266446 (3.2 MB)  TX bytes:1046654 (1.0 MB)       Interrupt:20 Memory:d3500000-d3520000 Am I going to have trouble flying with a reversed name on the ticket? I'm flying to Peru with my girlfriend next week and I realized that her name is incorrect on the reservation.  Instead of Sally Marie Jones it's Jones Marie Sally.  We're flying into Houston, then Lima, and in Lima we switch airlines for our flight to Cajamarca.  I booked the ticket with Orbitz who made the reservation with United. United booked the flight with LAN air, the Peruvian airline.  Both companies say that they are unable to change the name because of TSA restrictions not allowing a name change. I asked if I could just cancel and re-book, but United said they could not do this because there are multiple airlines involved.  A note has been attached to the reservation explaining the issue, and the United rep said that they would be able to print her boarding pass with the correct name. However, he said there are no guarantees that she'll be able to fly because although the boarding pass will be correct, the ticket will still have the reversed name.  I was prepared to just take the chance that it will all work out, seeing how it's really obvious the name is just backwards, but with recent events I'm worried that US airport security is going to be extra strict for the next little while.  Is this something I need to worry about, or are chances good that this will work out fine?  The United rep said there's a chance we could cancel the United flights and get a new ticket with the correct name for no cost, but we'd have to pay for new tickets within Peru ($500 out of the total $1300 ticket cost). I'd rather not do this unless there's a reasonable chance we'll have trouble at the airport. How to install packages in ubuntu without internet connection? Proving $\sin(x^2)$ Continuous but Not Uniformly Continuous with non-standard Definition Show that the bounded function $f(x) = \sin(x^2)$ is continuous, but not uniformly continuous, on the interval $[0,\infty)$.  The problem should be solved using the following definitions: Definition of Continuous Function Let $X$ and $Y$ be metric spaces. We sat that a function $f:X\rightarrow Y$ is continuous if given any open set $V \subseteq Y$, its inverse image $f^{-1}(V)$ is an open subset of $X$. Definition of Uniformly Continuous Function Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. If $E \subseteq X$, then we say that a function $f: X\rightarrow Y$ is uniformly continuous on $E$ if for every $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that for all $x,y \in E$ we have $$d_x(x,y) &lt; \delta \longrightarrow d_y(f(x),f(y)) &lt; \epsilon$$ For the continuity part it is clear that $f^{-1}(a,b), (a,b) \subseteq (-1,1) = \{x:f(x) \in (a,b)\}$ which appears to be an open set however I doubt this is sufficient as a proof since it holds for any function $f$. Not sure how to go about proving $f$ not uniformly continuous. Seems to be difficult to prove from the definition alone.