Recent content by grendel

i said a foreign keycan be repeated but it does not neccessarily have to be. it depends on the database. the only restriction is on the primary key that must be unique as its function is to identify a specific record. consider the rta database. the drivers license number in a "Driver Table"...

ERDs (Entity Relationship Diagrams) are just schemas. No difference at least in the IPT course

relational databases are not suited in implementing spatial (3d design, CAD)databases or handling image data. cannot deal with data that have complex interrelationships and difficulties may exist in processing data from within certain programming environments. thats all i could think of/find.

that looks good. in the LOANS Table Item Number and Student Number are referred to as foreign keys. While primary keys are unique, foreign keys can be repeated within a table. for example if a student has 4 loans then that students id will appear four times in the LOANS Table

No, Salary>55000 and Salary<50000 how can a persons salary be greater than 55000 and less than 50000???? no matching records.

Go here http://web1.fairfield-h.schools.nsw.edu.au/moodle/course/view.php?id=16 and scroll down to the word document called "normalisation". You will have to log in as guest. this is a limited time offer! Also you can check out this link...

pilot conversion. you could trial the new system in one of the offices over a period of time. depending on the result of the evaluation you could then deploy it in the remaining offices. it should be noted that you could argue a case for direct, parallel or phased.

analog sound can be represented as a wave (think sine wav). the different pitch and volume in the analog sound is represented by the wave's frequency and amplitude. the wave is sampled (readings of amplitude and frequency) and converted into a digital format such as wav, mp3, wma or...

permutations and combinations/probability question from hell! plus binomial theorem.

ok her goes. the total number of arrangements=(total with no restrictions)- ..........................[(total arrangements with all four letters adjacent)+ .............................(total arrangements with three of the letters adjacent)+ ...............................(total...

case 1: 3 different letters = 7 x 6 x 5 =210 (using letterbox technique) case 2: 2 E's = 6 x 3 (as the there are 6 choices for the third letter and it can be in one of three places) tf total = 228

Assuming you can only use each digit once: (i) 5! (ii) arrangements where they are together=6! x 2 arrangements where they are seperated by 1 digit=5 x 2 x 5! arrangements where they are seperated by 2 digits =4 x 2 x 5! tf total arrangements = 3600 (i) arrangements where e...

the answer in the back of fitzpatrick is wrong. had a look at the official solutions and the one i posted earlier is correct =)

assumption: probability that any particular word needs correcting is 1/800. tf p(more than one correction per page)=1-(p(0 corrections)+p(1 correction)) =1-(200C0 (799/800)^200 + 200C1 (1/800)(799/800)^199) = 0.026 which is not the answer...

T=2pi/n pi=2pi/n n=2pi/pi n=2 now, v^2=n^2(a^2-x^2) sub x=0, v=4 16=2^2(a^2-0) 16=4a^2 a^2=4 a=2 tf v^2=4(4-x^2) sub in x=1 v^2=4(4-1) v^2=12 v=sqrt(12)