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i wasnt trying to make you look like you dont know what you are talking about, i just said that it wasnt a protocol :s

CSMA/CD isnt a protocol its a method of handling contention used by the Ethernet Protocol

Most common stars are treated as buses and therefore use ethernet as well

Bus use Ethernet protocol

I thought this worked well for the SDD thread.. post questions you have and someone will answer them.. 1) Please don't post comments in this thread. 2) Questions and replies to questions only. 3) Please don't repeat your question if not answered. =)

it is the same amount of searches but it has to search through all 10,000 elements to find the next max value, where as an insertion only needs to shuffle it doesnt need to go through all 10,000 elements to find the max then go through all 9,999 to find the next max, etc etc

i worked it out so i talked about bubble in part i... then looked at part iii decided that insertion was the one that was the best so i talked about selection in part ii covering all 3 sorts in the section

http://www.boredofstudies.org/community/showthread.php?t=47165

I believe the correct answer is Insertion Sort but not for the reason you said... a Bubble sort and a Selection sort need to pass through all 10,000 elements each "pass" so if on avage it needs about 8,000 passes (assuming some elements are already in order) each sort would need to go through...

14 is not A the key is in the wording.. the first row is the year.. year 1, 2, 3 the second row is the ages 10,11 etc and the 3rd roow is how many people in that year are that age... so in year 1 there are 12 people of age 10 and in year 2 there are 9 people of age 10 so the answer is (0,2) + (2,2)

9 2 for 21 2 for 22 2 for 23 3 for 25 They should give 8 page booklets... 4 was ridiculous most of mine filled all booklets except 25 which was two pages in the third...

don't beat yourself up about it.. the question was worth 6 marks chances are you wont do everything perfect if you didnt validate it (between 1 and 10) you might lose half a mark and another half for something else.. but if you did validate it and didnt do something else you will still get 5 /6

you were supposed to write one from scratch but it was easier to take parts of their algorithm.. because it did most of the job if it works, it works, regardless of whether you used parts of the code in the earlier question or not

i did: Begin print_adjacent unsold --Input "How many adject seats" ; amtadj --IF amtadj > 0 AND amtadj < 11 ----row = 1 ----WHILE row <= 8 ------Seat = 1 ------WHILE seat <= 10 --------available = 1 --------IF array(row,seat) = "unsold" THEN ----------FOR i = seat to (seat + amtadj)...

given the relative ease of this test i would say an 85 - 88 raw mark would be needed to get a band 6

The question was so easy.. i mean converting a number from binary to decimal then back to binary was a give away 4 marks.. and identifying the AND and NAND gates was easy too.. how did people "compare and contrast" them.. that threw me off a bit but overall the option was a giveaway 10-14 marks...

nice work pop'n'fresh!!

1: C 2: A 3: C 4: D....? (maybe B) 5: D 6: A 7: D 8: D 9: A 10: C 11: C 12: A 13: C 14: B 15: B 16: B 17: B 18: D 19: C.... (Maybe D) 20: B

yes you are right.. it was a trick... the test was easy compared with previous years

I think the key is "end-user" not the whole company.. the end-user only really has a say on how much he/she can do therefore how much training they need