Computer Forums Little challenge.. (You'll prob find easy)
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 05-30-2005, 02:06 PM #11 Site Team     Join Date: Dec 2002 Posts: 15,233 Re: Little challenge.. (You'll prob find easy) 2^24? 16,777,216 bits __________________ __________________ [url=http://www.LNXPS.NET]LNXPS.NET - The XPS Library]
05-30-2005, 02:12 PM   #12
Wizard of Wires

Join Date: Feb 2005
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Re: Little challenge.. (You'll prob find easy)

Quote:
 Originally Posted by David Lindon 2^24? 16,777,216 bits

So how many bytes based on the model in the question?
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 05-30-2005, 03:22 PM #13 In Runtime     Join Date: May 2005 Posts: 195 Re: Little challenge.. (You'll prob find easy) Arrhh! I have to take this exam next year, i hope most questions are easier than this
05-30-2005, 03:47 PM   #14
Daemon Poster

Join Date: Jun 2004
Posts: 838
Re: Little challenge.. (You'll prob find easy)

Quote:
 Originally Posted by David Lindon 2^24? 16,777,216 bits

on the right lines! follow it through

well kinda, 2^24 would mean there space for 2MB of RAM. lol

just on the side- you picked 2^24, this gives 16.7 mil, number of colours in a JPEG image, jpegs are 24bit colour.
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05-30-2005, 04:15 PM   #15
Fully Optimized

Join Date: May 2005
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Re: Little challenge.. (You'll prob find easy)

Quote:
 Originally Posted by bigbondfan Arrhh! I have to take this exam next year, i hope most questions are easier than this
I'm taking it as well... heh.

05-30-2005, 04:25 PM   #16
Daemon Poster

Join Date: Jun 2004
Posts: 838
Re: Little challenge.. (You'll prob find easy)

Quote:
 Originally Posted by Bahawolf I'm taking it as well... heh.
dont worry, most of the questions are a easier!

Eg, no7 section I - what piece of hardware must be present in order to network their stand alone computers? (shortened version of Q)
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 05-31-2005, 05:03 AM #17 Daemon Poster     Join Date: Jun 2004 Posts: 838 Re: Little challenge.. (You'll prob find easy) ok, the solution (well what I got) David was along the right lines, 2^32 gives the number of addressable storage location in memory. 32 though david not 24, as 32 is the address bus. Now each storage locations holds one word. A word being the amount of data a processor can process in a single operation. We are used to 32 and 64, but in this question its 24. The word length is the number of data line son the data bus. So if each storage locations holds 24bits we have a computer that can hold ((2^32)x24) bits of Ram. ((2^32)x24) = 1.030792151x10^11 (calcualtor puts it into standard form) thats bits so divide by 8 to get bytes 1.030792151x10^11 / 8 = 1.288490189x10^10 bytes to get kilobytes divide by 1024 1.288490189x10^10 / 1024 = 12582912 KB by 1024 again 12582912 KB / 1024 = 12288 MB and by 1024 again gives 12288 MB /1024 = 12 GB __________________ --- Every man dies, not every man really lives ---
 05-31-2005, 09:43 AM #18 Golden Master     Join Date: Dec 2004 Posts: 12,208 Re: Little challenge.. (You'll prob find easy) Whew, I'm going to paste that in Notepad to review later. That's good info. It's mainly just general conversion from there: ((2^32)x24) bits __________________ *Fact: Microsoft Window's Blue Screen of Death vs Computerforums.org's White Screen of Death. Which is worse?
05-31-2005, 11:03 AM   #19
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Re: Little challenge.. (You'll prob find easy)

Quote:
 Originally Posted by jay8990 Hi so I just sat my Higher Computing Exam.
I just did the Advanced Higher one... got an A at Higher last year.
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 05-31-2005, 01:05 PM #20 Daemon Poster     Join Date: Jun 2004 Posts: 838 Re: Little challenge.. (You'll prob find easy) cool, i was going to stay on and do the AH, but hated school. lol might go do it at college in my year out, and AH Physics. Really enjoy physics as well. __________________ __________________ --- Every man dies, not every man really lives ---