IPT Question Thread + Useful Info

[Hua Fei]

Question:With out looking at answers, answer a) and c) as b) is easy.
(Image removed from quote.)

Btw where is this question from? And do you have answers?

[aimbotted]

Btw where is this question from? And do you have answers?

ive seen them before, they're from a hsc past paper i think one of the 2011-2015 ones

[BPunjabi]

Binary code? Do you mean you don't understand how to turn a decimal to a binary?

I dont get changing the value at the end... look at every multiple choice Question, its usually the last.

[BPunjabi]

No worries! Glad it helped!   
Also, the answers to your questions:

a) Cyclic redundancy check (CRC) is a method used to detect errors in transmission – a method 99.99% more effective than check sum. Hence, in this information system, it could be used when the details of the customer when they are applying for a loyalty card are transmitted to the department store.
Note: I don’t think I wrote an adequate amount…and if I did this correctly….

c) In this information system:
- Internet would be used to access the online department store for shopping and applying for a loyalty card. It would also be used when the customers receive advertising emails from the business partners of the online department store.
- Intranet would be used when the store shares information with its employees.
- Extranet would be used for when the store shares information with its business partners. 
Note: Not too sure about this question either……     :’(

Nice Job with c) but wouldnt you say with a) the process behind CRC?

[Hua Fei]

I dont get changing the value at the end... look at every multiple choice Question, its usually the last.

How to convert decimal to binary (I think that's your question?):
For example, how is 72 represented as a binary?
Answer: 1001000 (working out in the picture attached)

Hope this answered your question!

[BPunjabi]

How to convert decimal to binary (I think that's your question?):
For example, how is 72 represented as a binary?
Answer: 1001000 (working out in the picture attached)

Hope this answered your question!

LoL bro, I know how to do binary ahahah and you screwed up your method.. its 0-1-2-4-8-16-32-64-128
I mean a question that says like pick the wrong message. Shows different codes. Its either odd even and you have to pick which code it is

[aimbotted]

hes talking about parity bits, which are 001011 0 with the parity set to odd or even.

[BPunjabi]

hes talking about parity bits, which are 001011 0 with the parity set to odd or even.

Yep

[Hua Fei]

hes talking about parity bits, which are 001011 0 with the parity set to odd or even.

Yep

Ah, right.
Well then,
For parity checks, you choose to transmit the data with an odd parity bit or an even one.

E.g. the binary to be transmitted is 0110101 and you choose odd parity. Choosing odd parity means that the number of 1's must be odd - if it's not, then you make it odd by making the parity bit 1. However, if it is, then the parity bit would be 0 in order to leave the number of 1's odd.
In this case, the number of 1's in the binary are 4. We have chosen odd parity hence the parity bit has to be 1 to make the number of 1's odd. As a result, the transmitting data would read as: 0110101 1

On the other hand, if even parity was chosen, the parity bit would be zero and the data to be transmitted would read as: 0110101 0

Hope this helped!

[Hua Fei]

LoL bro, I know how to do binary ahahah and you screwed up your method.. its 0-1-2-4-8-16-32-64-128
I mean a question that says like pick the wrong message. Shows different codes. Its either odd even and you have to pick which code it is

Note: the "0-1-2-4-8-16-32-64-128" method is only used for converting binary to decimal not decimal to binary. I did the method where you convert decimal to binary (i.e. 72 is converted to 1001000)

[BPunjabi]

Ah, right.
Well then,
For parity checks, you choose to transmit the data with an odd parity bit or an even one.

E.g. the binary to be transmitted is 0110101 and you choose odd parity. Choosing odd parity means that the number of 1's must be odd - if it's not, then you make it odd by making the parity bit 1. However, if it is, then the parity bit would be 0 in order to leave the number of 1's odd.
In this case, the number of 1's in the binary are 4. We have chosen odd parity hence the parity bit has to be 1 to make the number of 1's odd. As a result, the transmitting data would read as: 0110101 1

On the other hand, if even parity was chosen, the parity bit would be zero and the data to be transmitted would read as: 0110101 0

Hope this helped!

Ah Cool thanks !!!!!

- So how would you answer this question?

[aimbotted]

Answer's B.

Here's why, the first bye there 3 0's and 5 1's this makes it an odd parity as the parity bit is 1 to make the 1's odd. Whilst in the 2nd and 3rd byte there are even amounts of 1's and 0's. As state before the first byte's parity is ODD, whilst the second byte and third byte have even parities. Therefor the Parity is odd and the corrupted data bytes are second and third

[BPunjabi]

Answer's B.

Here's why, the first bye there 3 0's and 5 1's this makes it an odd parity as the parity bit is 1 to make the 1's odd. Whilst in the 2nd and 3rd byte there are even amounts of 1's and 0's. As state before the first byte's parity is ODD, whilst the second byte and third byte have even parities. Therefor the Parity is odd and the corrupted data bytes are second and third

OMFG, I love you!!! Thank you so much... I get it now!!!! do you have a similar question that you could ask me on this binary thing and I answer it and you tell me if Im right?

[Hua Fei]

OMFG, I love you!!! Thank you so much... I get it now!!!! do you have a similar question that you could ask me on this binary thing and I answer it and you tell me if Im right?

Have a go at these two     

[BPunjabi]

Have a go at these two     

Im guessing a) for 1, I dont know its odd and a 1, all the rest are even