I am transcribing a naturalistic language sample during a board game activity.  There is lots of rote counting when taking a turn, which makes up a good portion of my language sample.  I am hesitant to include the rote counting as part of the MLU because this is not truly "language" just memorization, and will inflate the MLU sample I think. Shoudl I not count rote counting at all, and just exclude, or count it as 1 because it is one utterance.?  Thanks!
211150
Assuming you are not doing research and you are trying to identify a specific skill. You don't need to worry too much. You should use your clinical judgment.  If you are trying to decide if the child is in need of services then you have many tools you can use to help make this decision.  If you are trying to get a baseline then adding counting will deflate your sample, not inflate it.  It may be more important for you to decide if the counting was appropriate?
473394
Hi I work as an online instructor, and previously at Berlitz. They use the "natural method of teaching" where you teach and speak only the targeted language. I left about a year ago so I don't really their approach to online teaching.\n\nHowever, when I left I went into online training and ever since I've developed my own methodology and the only online esl tutoring info website. I went into online training and I've since developed several teaching materials based on the natural method (not the Berlitz method) for teaching ESL. This approach seems to work well, and I've been using it online (and inside the classroom) for about a year using Skype's chat and voice conferencing feature - sometimes 1:1 or 5:1. I use interactive video, questioning techniques .... which are all accessed by students on the web as they speak to me. You can also teach grammar, writing, and just about any other subject but this would be a slightly different approach.
521422
So is this sequence based purely on a mathematical function?\n\nOkay, so there's a 52 in there, could this sequence somehow relate to the calendar?\n\nWhat sort of scientist? I'm a computer scientist and nothing's glaringly obvious about those numbers.\n\n** GOT IT *** 68! Er or Erbium, the sequence spells out Brainteaser based on the atomic weights of B, Ra, In, Te, As, and Er.
491060
Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.
443236
If you have access to a copy of the ARRL handbook, there are several nomographs in there for RF calculations.  Not exactly a slide rule, but it will do the job.  I copied the graphs and laminated them.  Then I use a dry erase marker to line up the data points and read the results.
654955
In my opinion, it doesn't take that large of a sample in order for the CLT to take effect.  Although in many classes, it is said that a sample larger than or equal to 25 is needed, that's not always necessary.\n\nIt does really depend on the shape of the population, though.  If the population is roughly symmetric, then a sample of 10 or possibly less might be sufficient.  If not, if the population is skewed and/or has outliers, then a larger sample would be necessary.\n\nAn elemetary stats book by Weiss has some good guidelines.  I don't really remember what exactly they are, but, basically, you should look at the histogram and the normal qq plot of the of the data.  Look at the histogram is roughly symmetric, and the normal qq plot is roughly linear.\n\nOne of my instructors once usually looked at the estimates for the skewness and  the kurtosis to make the determination.  I think that if the skewness is less than 2 in absolute value and kurtosis is less than 2.5, then the t test would be OK.\n\nedit: (This is the email I sent you earlier)\nI said the skewness and the kurtosis is less than those numbers, not greater than.\n\nYou can use Jarque-Bera as well as looking at the skewness and kurtosis.\n\nAlso, when I say kurtosis, I mean that it is the regular kurtosis minus 3, since 3 is the kurtosis of the normal.\n\nIn terms of what is considered minimum, well, the absolute minimum number of values you need is 2 of course.  But again, that's only if the population is considered normal.\n\nThe CLT converges much faster in some populations versus others.  So sometimes 10 is enough, sometimes not.  That's why I didn't set a particular number.\n\nedit: Yes, nonparametric tests can be used for pretty much any sample size.  But again, the sample size in which you transfer from parametric to nonparametric should depnd on the shape of the population.\n\nAlso, alyxsylvr, you would know that standard errors typically do not depend on the sample size as compared to the population size; it depends on the absolute size of the sample.
435355
Two words:\nStill Counting
379488
Dont worry , I know the book that has done it for you.\nthat is :"Compilers: Principles, Techniques, and Tools. Alfred V.                    Aho, Ravi Sethi, Jeffrey D.Ullman"\nif you could not reach that book search its slides on the web or refer to Jeffry D.Ullman web site to get lecture notes at: http://infolab.stanford.edu/~ullman/\nit has a complet source code for a lexical analyzer for a sample grammar.\nat first step you must write a garmmar that has no left recortion\nand for any nonterminal write a function that calls other nonterminals or matches terminals.
342063
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
461933
I thought the rule was two of any level entry...I'll do a little research and see what I can turn up.\n\nLater: as I thought.  One website on the topic says "There must never be a lone subtopic. You cannot divide something into less than two parts; therefore, every A must have a B; every 1 must have a 2. "  See below for the website reference.  However, since Roman numerals indicate major topics, can you really write a decent paper that only has two main points/topics?
524522
Ah, LSAT prep.  This one appears to have two interchangeable answers, so I'll show both.  Here're my steps:\n\n1. -------------------- (There's my rectabgular table)\n\n2. MM------------------MF\n\n          (xy) (xx)(xy)(xx)\n3  MM-----------------MF\n         (xy) (xx) (xy)(xx)\n\n         (xy) (xx) (xy) (M/S)\n4 MM------------------MF\n         (xy) (xx) (xy) (M/S)\n\n#5 is where it gets ambiguous, since we don't know what side Maria's on.  I'll start on one side and switch her if necessary\n         (xy) (xx) MrL M\n5 MM------------------MF\n         (xy) (xx) (xy) S\n\n          (xy) GM MrL M\n6 MM------------------MF\n          (xy) MrsL GF S\n\n          B GM MrL M\n7 MM------------------MF\n          U MrsL GF S\n\nThis is a correct answer, but it turns out that had I placed Maria on the bottom and her sister on the top, I would've ended up with a different answer that satisfies the same criteria:\n       B   MrsL GF  S\nMM--------------------MF\n       U   GM  MrL M
240586
a) That means the number has to start with 5\nso you only need to choose the second and third digits.\nThere are 4 choices for the second and 3 choices for the third\n4*3 = 12\nb) The first digit must be 1 or 2\n2 choices for the first 1\n4 choices for the second \n3 choices for the third\n2*4*3 = 24\nEDIT: I read the part b incorrectly. This is the answer if no repetitions are allowed.\nIf repetitions are allowed, then you still have 5 choices for the second and third digits.\n2*5*5 = 50
261507
Yep thats right, it is in 7/8, but after the solo it changes to 4/4 and then back to 7/8\n\nPS...Just have to say it - ebayjunckie, what a preachy arrogant little tosser, why don't you save your didactic ramblings for some other place? I have a degree in English and Linguistics, but I wouldn't dream of lecturing someone on their use of English, language is for COMMUNICATION - for many people on here, English is not their first language, and I think they do damn well on the whole to make themselves understood, there is no need for a preachy little sod like you, esp as you are probably a Yank, so in no position to teach someone to talk ENGLISH\n\nPPS....what the hell does your inane rant have to do with the question anyway????
509409
You put a round bracket "()" on the number line when the sequence doesn't include the number where that bracket goes. For example, if the final answer you get is "all the numbers between 2 and 8, excluding 2 and 8" then you use the round brackets. You'll know your answer excludes a number or numbers because it will be "bigger than" or "smaller than" only, not "bigger than or equal to" or "smaller than or equal to", in other words, ther will be no line underneath the inequality sign. \nNow, if you do get something like "x is bigger than or equal to  8 and smaller than or equal to 2" than you know your answer includes 2 and 8, so you use the brick bracket, "[ ]". Good luck.
212648
"Yep, it was an artificial language which 'died'."  Hahaha\n\nMi parolas esperanton! (I speak esperanto) \n\nI speak esperanto better than english :P\n\nI have a lot of friends Esperantists!!  \n\n"Its phonetics is not very simple, it is a sexist language that considers for example, that a woman is just a female version of a man, and its derivational processes are not always very clear."\nThat is very subjective!  \n\nIt is more difficult to find idists than Esperantists!\n\nIn my opinion, what is important is to diffuse the "internal idea" (a neutral international language), and is better if is done from a sufficient strong movement(the esperantist).  Supporting Ido just would weaken a little the esperantist movement, but would not help to diffuse "the internal idea".\n\n"I've heard of it. I had a friend on livejournal.com who was studying it."\n\nSee:\nhttp://www.livejournal.com/interests.bml?int=esperanto\nhttp://community.livejournal.com/esperanto/  \n\n;D
639803
For me, it's 12. Why?\n\nOh no, this is one of your asl tricks, isn't it?
159712
Q: What do you call someone who speaks two languages?\nA: Bilingual.\n\nQ: What do you call someone who speak one language?\nA: American.\n\nIn many parts of the world, it's common to speak two or three languages.  Many places speak a European language as a second to their own, or a "lingua franca" for their region, such as Arabic (you can hear it from Morocco to Afghanistan) or Swahili (20+ African countries and 50+ million people).  \n\nThe Europeans, North Africans and Middle Easterners speak several languages because they are exposed to them regularly.  You'll find many Scandinavian and Baltic (Estonian, Latvian and Lithuanian) athletes who compete at the world level speak English because they receive English TV transmissions.  (The Baltic states listened to Finnish transmissions illegally.  The Finns didn't mind, but Moscow did!)\n\nFortunately, most who do listen to English broadcasts listen to BBC, and NOT to Mouthpiece of America - oops, Voice - VOA.  Sad to say, it's the English speakers who are the biggest morons and don't speak more than our own tongue.  Not taking my French classes seriously in elementary and high school was the second biggest mistake of my life.  (Listening to my parents was #1.)\n\nAnd to see idiot reichwingers in the US who think Americans speaking Spanish is a problem, I can only shake my head.  South America is 500+ million in size, is a growing economic power, they speak Spanish (except for Brazil), and the US wants to ban Spanish in schools and government.  Incredible. 9_9