 | | Re: math question (or pattern... whatever...)
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In lugnet.off-topic.geek, Larry Pieniazek writes: Another site I happened to stumble across (which is so chock full of math gadgets and java applets that I added it to the header...): (URL) this writeup of the problem, along with a simulator for you (...) (22 years ago, 2-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) Ouch. I haven't had those Win2K horror stories. I got a mouse that hangs, and really slow access to the file system. I need to re-format and reinstall on this Dell Inspiron 8100, and upgrade the desktop. (...) I like 2k a lot better. I just (...) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) MS burns me to no end-- Win2K on Toshiba 3000 laptop--netmeeting causes BSOD! Beyond that, I finally get laptop just the way I want it (Diablo II, EQ, and Mindstorms/NQC, as well as business apps...) and the DVD Video overlay gets corrupted. (...) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) I'm not too happy with the current installation of Win2K I'm running. But, I'm a bit busy to do something about that right now :( Once I no longer need the secondary machine for rendering (got a feeling that will be never, but I'll have to (...) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) I thought maybe you were having a bad MS day ;-) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) I know what day it is, read what I wrote at the bottom of the post :-) -Tim (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Re: Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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(...) What day is it? (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | Judge Kotelly Admits Error, Will Create New Ruling with Judge Jackson
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Just read this.... wow. (URL) from the page above) April 1, 2003 BROADCAST TRANSCRIPT: "U.S. District Judge Colleen Kollar-Kotelly, who oversaw the settlement between Microsoft and the U.S. Department of Justice, issued a formal statement to the (...) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek)
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| | serious vulnerability present. all doomed. over.
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This was funny, coming from a well respected moderated vulnerability list :) FUT to off-topic.debate, if you want to talk about the contents :) ----- Forwarded message from "Security Experts, Liability Limited" <throwaway@dione.ids.pl> ----- (...) (22 years ago, 1-Apr-03, to lugnet.off-topic.geek, lugnet.off-topic.debate)
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| | Re: On a scale of 0 to 10?
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(...) I don't see an easy way to do this. I tried substituting x=e^B Then I get a polynomial of degree 5. Only the positive roots lead to real values of B. (since e^any real number > 0) This is solvable, but very difficult to do by hand, especially (...) (22 years ago, 28-Mar-03, to lugnet.off-topic.geek)
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| | Re: Geek Hierarchy
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(...) Hey, where are the toy-collecting geeks on this thing? I feel so underrepresented! :) Hilarious. -jeremiah- (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
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| | Geek Hierarchy
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Posted without further comment... (URL) (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
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| | Retina scanner?
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You do realise that that 'retina' scanner on CLSOTW is actually just photographing the cornea, not the retina, don't you? Jason Railton (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) aHA! Much oblige! I now get (approx): s = 5.39 * e^(2.625q) - 0.39 Whew. Of course, now here's a totally different question. In order to get that point, I cheated. I couldn't solve: e^(B/4) + e^(-B) = 2 using algebra, but using other means, I (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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Hi all. The reason the equation is unsolvable is because B is an unneeded constant. Why? A*e^(q+B)=A*(e^q)*(e^B) It is impossible to distinquish between A & e^B. What you need to solve is an equation of the form s=A*exp(B*q) + C This is the form of (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) Oh yeah! (...) Oh... yeah. (...) Hm. Double checked the math-- it appears solid [1], which would mean that there's no viable solution for: s = Ae^(q + B) + C for coordinates (-1,0), (0,5), (0.25,10). Darn. Hm. I guess that also means that any (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) *snip* (...) summed exponents like that can be re-written in this way: e^B * e^.25 + e^B * e^-1 = 2e^B (e^.25 + e^-1) * e^B = 2e^B Uh oh. Divide both sides by e^B and we've got (e^.25 + e^-1) = 2 1.652 = 2 Which ain't right. I think your math (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) How about a double-linear solution? from -1 to 0, translate linearly from 0 to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a single equation, but that seems to accomplish what you want, right? Adrian (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | On a scale of 0 to 10?
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I've got a function that represents a quality rating, based on 2 parameters, which are on a scale of 0 to 1: (d-k)*k => quality (q) Notice that this yields a sort of wacky result, insofar as the quality can vary between -1 and 0.25, not -1 and 1 or (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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