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Instructions for Side by Side Printing
  1. Print the notecards
  2. Fold each page in half along the solid vertical line
  3. Cut out the notecards by cutting along each horizontal dotted line
  4. Optional: Glue, tape or staple the ends of each notecard together
  1. Verify Front of pages is selected for Viewing and print the front of the notecards
  2. Select Back of pages for Viewing and print the back of the notecards
    NOTE: Since the back of the pages are printed in reverse order (last page is printed first), keep the pages in the same order as they were after Step 1. Also, be sure to feed the pages in the same direction as you did in Step 1.
  3. Cut out the notecards by cutting along each horizontal and vertical dotted line
To print: Ctrl+PPrint as a list

18 notecards = 5 pages (4 cards per page)

Viewing:

Computing

front 1

What is encryption?

back 1

Converting plaintext(readable data) to ciphered text (unreadable)

front 2

What is readable data

back 2

plaintext (normal text)

front 3

What is ciphered data

back 3

unreadable data

front 4

What is parity bit

back 4

A bit(0s or 1s) added to a byte to ensure total number of 1s is even for even parity check and total number of 1s is odd for odd parity check

front 5

Why is parity checking important

back 5

To ensure data was received correctly.

front 6

Define strong password policies.

back 6

Must use letters, numbers & symbols
Must be more than 10 characters
Avoid using names or simple words
Change passwords regularly

front 7

What is biometric security

back 7

Security that uses fingerprints, facial recognition, or iris scans

front 8

Explain the purpose of 2FA

back 8

Add an extra layer of security

front 9

❓ Question:

You received the following 8-bit data packet:

CopyEdit11010111

Assume that parity is EVEN.

back 9

Yes, because we are checking for even parity and the total number of 1s is even.

front 10

❓ Question:

You received the following 8-bit data packet:

CopyEdit10110010

Assume that parity is ODD.

back 10

No, because we are checking for odd parity check but there is even total number of 1s when its supposed for the total number of 1s to be odd.

front 11

❓ Question:

You are given this 7-bit data packet:

1010110

Use even parity check

back 11

The parity bit should be 0 ( in exam you can put the parity bit in the parity bit row one!)

front 12

❓ Question:

You are given this 6-bit data packet:

110101

Use odd parity.

back 12

1,0 / 0,1 (any order is fine but the total number of ones but be odd at the 8 bit(6 bit with the extra 2 parity bit) total number of ones.)

front 13

READ THIS/ FLIP THE CARD

back 13

The exam will ask this question

Identify if data transmission from data packet is successful by even/odd parity

This means check if the total num of ones match which parity check you are checking.

front 14

Describe how to locate error in 2d Parity

back 14

First check each row and column and check which parity check it is (even or odd)
If both column and row is wrong, find where a row and column that have an issue intersect and flip that bit to fix the error.

front 15

Can parity checking fix errors?

back 15

So the data wont be sent incorrectly or

front 16

Give real world example of 2FA

back 16

(Learner's Opinion)
E.g. When you login to google by email and password and google sends a code to the user's email to confirm and the user must enter the code to access the account this provides two layers of security.

front 17

Compare passwords and biometrics for security. Which is more reliable?

back 17

This question both is correct but you need a good explanation to

Passwords is more reliable rather than biometrics, because people can use a picture of your face or your fingerprint to access your account but for passwords you can put a very strong password and hackers will not be able to use brute force attacks to guess your password.

front 18

One limitation of parity checking, give one example

back 18

It can only detect odd number of errors not even number of errors, for example a data packet 11011011 has two bits changed during transmissions, 10111011, the it still has the same total number of 1s in the data packet and so the wrong data will be transmitted.