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Old 05-30-2005, 03:06 PM   #11
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Default Re: Little challenge.. (You'll prob find easy)

2^24?

16,777,216 bits
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Old 05-30-2005, 03:12 PM   #12
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Default 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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Old 05-30-2005, 04:22 PM   #13
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Default 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
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Old 05-30-2005, 04:47 PM   #14
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Default 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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Old 05-30-2005, 05:15 PM   #15
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Default Re: Little challenge.. (You'll prob find easy)

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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.
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Old 05-30-2005, 05:25 PM   #16
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Default Re: Little challenge.. (You'll prob find easy)

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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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Old 05-31-2005, 06:03 AM   #17
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Default 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
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Old 05-31-2005, 10:43 AM   #18
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Default 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
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Old 05-31-2005, 12:03 PM   #19
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Default 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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Old 05-31-2005, 02:05 PM   #20
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Default 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.
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