Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
stuck on a question...
In a survey of 2400 internet users, 95% said they had functioning anti-virus software on their computer. In order to confirm this, the computers (those belonging to the 95% who had made the claim) were checked. It was found that only 75% of these computers had functioning anti-virus software. How many of the 2400 people surveyed had functioning anti-virus software on their computer?
|
corsamike
Member
Registered: 26th Sep 01
Location: WIGAN Drives: Clio Sport 182 ; ZX 250R
User status: Offline
|
1710
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
Cheers how about this one?
If 5 users are connected to a wireless network that offers a maximum data rate of 54 Mbps and a throughput of 60%, which of the values given below is the average practical message data rate available to each user?
|
John
Member
Registered: 30th Jun 03
User status: Offline
|
Is this primary 3 homework?
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
54 * 0.6 / 5 = 6.48mbps
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
quote:
Originally posted by John
Is this primary 3 homework?
What I was thinking tbh...
|
corsamike
Member
Registered: 26th Sep 01
Location: WIGAN Drives: Clio Sport 182 ; ZX 250R
User status: Offline
|
sorry failed to show working out earlier so for full marks.........
2400 * 0.95 = 2280
2280 * 0.75 = 1710
lol
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
Just wanting some help as im not that great with percentages.
A computer is connected to the internet via a 9.6 Mbps broadband connection. Data is received at an average rate of 600 kbps when navigating in a web-based ‘virtual world’. What percentage of the available bandwidth is being used (to 2 significant figures)?
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
 3 years of accounting degree taught me to always show workings, you get more marks for showing them than actually getting the answer right, all about the method marks  
|
GF-91
Banned
Registered: 8th Jul 09
Location: Burnley!
User status: Offline
|
Are you doing your childs homework? :/ Tell them to GTF and do it themselves.
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
600/9600 * 100 = 6.25%
edit - I think
[Edited on 23-12-2009 by dannymccann]
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
Think your right danny I got 6.3% so round it up I guess
Last two Im stuck on:
A clip of an animated film is 15 seconds long. The file containing the film is 9.6 MB. On average, how much data (to 3 significant figures) does each second of the film use?
[Edited on 23-12-2009 by Dione J]
|
corsamike
Member
Registered: 26th Sep 01
Location: WIGAN Drives: Clio Sport 182 ; ZX 250R
User status: Offline
|
i used the "post quick reply" box and therefore gave a quick answer!! LOL
but i agree, its all in the method........getting the answer right at the the end of it all does help though!! 
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
9600 / 15 = 640kb
Depends if you are using layman mb's or 1337 mb's though
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
Options I have is:
A 125 B D 370 KB
B 890 B E 655 KB
C 123 KB F 1.41 MB
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
Well I would say your answer is E then
|
Ian
Site Administrator
Registered: 28th Aug 99
Location: Liverpool
User status: Offline
|
9.6 x 1024 = 9830.4
9830.4 / 15 = 655.35
I'm guessing E.
Tell me, when you say 'stuck', do you mean tried a few things and spent some time on it or read it, couldn't even start it, and posted on here?
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
Thats the figure I was grabbing for Ian, 1024kb 
|
Cosmo
Member
Registered: 29th Mar 01
Location: Im the real one!
User status: Offline
|
Is this some sort of homework for your young cousin or something?
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
quote:
Originally posted by Ian
9.6 x 1024 = 9830.4
9830.4 / 15 = 655.35
I'm guessing E.
Tell me, when you say 'stuck', do you mean tried a few things and spent some time on it or read it, couldn't even start it, and posted on here?
Thanks Ian.
I have tried but got so many different answers just confused myself :O(
|
Ian
Site Administrator
Registered: 28th Aug 99
Location: Liverpool
User status: Offline
|
Post your workings in here. The only way you'll learn to tackle a question like that is to consider why each of your approaches are not correct.
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
In an audio conferencing system, each person’s speech is sampled 4000 times per second and 256 quantisation levels are used for each sample. When one person speaks for 3 seconds, how much data (to 3 significant figures) is sent?
4000*3= 12000? (11.7kb)
A 165 B D 1.09 KB
B 596 B E 3.25 KB
C 712 B F 11.7 KB
|
dannymccann
Member
Registered: 9th Aug 06
Location: Doddington, Lincolnshire
User status: Offline
|
Well thats F but nowhere in that question does it say how big each sample is? You are simply assuming 4000 samples = 4000b....
|
Ian
Site Administrator
Registered: 28th Aug 99
Location: Liverpool
User status: Offline
|
4000 x 3 is the number of samples, not the data rate. I think in this case, F is not the answer.
Do you have any more information anywhere on how to deal with the 256 quantisation levels?
|
Dione J
Member
Registered: 22nd Sep 04
Location: West Midlands Drives: Leon Cupra Turbo
User status: Offline
|
Only info i can find is:
Each sample must be allocated a number which represents how loud the sound is at that time. This number is given to a certain precision (a bit like rounding up or down to the nearest whole number), and you need to decide how many levels of precision you will use. These are often called quantisation levels, and the process of allocating each sample to one of these levels is called quantisation.
In order to make a good digital representation of the original sound you need to take very many samples per second. For high-quality digital audio you also need a large number of quantisation levels. So, the higher the sampling rate and the number of quantisation levels, the better the quality and the larger the amount of data.
|
