 | | 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 (21 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? (21 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 (...) (21 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> ----- (...) (21 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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