Upcoming 35s question #2 - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: Upcoming 35s question #2 (/thread-118144.html) |
Upcoming 35s question #2 - Gene Wright - 07-11-2007 The answer to question 1 was E. The early models did not have a lowered-ON key.
No multiple choice this time.
Re: Upcoming 35s question #2 - Thomas Radtke - 07-12-2007 Let me say, as much as can be represented by floating point numbers within the limits of the machine :^).
(I was right on q#1, I can afford to fail this time :-) Edited: 12 July 2007, 7:19 a.m.
Re: Upcoming 35s question #2 - Gene Wright - 07-12-2007 Answer is well over 2400 floating point, full precision numbers.
Re: Upcoming 35s question #2 - Paul Dale - 07-12-2007 That isn't exactly a precise answer to the question...
Re: Upcoming 35s question #2 - Gene Wright - 07-12-2007 Lol. Ok, 2514 is the actual, non-useful number. 801 x 3 + 26 x 3 + 6 x 3 + 5 x 3 which fills up all 801 indirect registers, all 26 lettered registers, all 6 stat registers (which have to be stored indirectly) and all 4 stack registers and lastx too. Reality is that you can now have indirect registers 1 through 800 filled with 3-D vectors containing 3 full precision real numbers using the program supplied in the learning module.
That ought to do for now. :-)
Re: Upcoming 35s question #2 - Thomas Radtke - 07-12-2007 Indeed, my answer was way better. You just have to read the question the right way ;-).
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