Convert these numbers from Decimal to Binary 111: 66: 252 11 20 Convert these numbers from...

Convert these numbers from Decimal to Binary
1. 111:
2. 66:
3. 252
4. 11
5. 20
Convert these numbers from Binary to Decimal
1. 00110110
2. 11111000
3. 00000111
4. 10101010
What is the Default Subnet mask of Class A IPV4
What is the Default Subnet mask of Class B IPV4
What is the Default Subnet mask of Class C IPV4
What is the CIDR notation or / short handwriting of Subnet masks:
1. Class A: /?. Explain the reason
2. Class B: /? Explain the reason
3. Class C: /? Explain the reason
Identify which IPV4 class these IP belong to
1. 6.0.0.0 Class?
2. 120.0.0.0 Class?
3. 130.30.0.0 Class?
4. 199.100.1.0 Class?

Please choose the correct answer and add any additional information. Thanks :)

Solutions

Expert Solution

Hi, I would like to answer you with these questions , I hope you will get something new to learn and if so then please give the thums up to the answer and not wasting much time lets head towards the answers:

Decimal to Binary

111 - 1101111
66 - 111110
252 - 11111100
11 - 1011
20 - 10100

Divide the number by 2 and you need to line up the remainder to get its binary

Binary to Decimal

00110110 - 54
11111000 - 248
00000111 - 7
10101010 - 170

The binary number is calculated as first digit is 2⁰ then increasing 0 to 1 then 2 and so on and calculte the sum at last like in 00000111 2⁰+2¹+2² = 1+2+4 = 7

Default Subnet Masks and their CIDR

Class A - 255.0.0.0 /8
Class B - 255.255.0.0 /16
Class C - 255.255.255.0 /24

The information which is used is that the 255 used block contains the network bits in the ip address rest with zero are host bit so Class A is like : 11111111:00000000:00000000:00000000 (N:H:H:H)

Identify the class of ip address

6.0.0.0 - Class A
120.0.0.0 - Class A
130.0.0.0 - Class B
199.100.1.0 - Class C

Ranges of ip addresses are as - Class A 0-127 ( face value ) Class B 128-191 Class C 192-223 After this there are for different other aspects Class D and E research and government uses.

So this was the explanation with the answers, Please give thums up to the answer and do comment for any queries

Thanks and Happy to help :)

HAPPY LEARNING

venereology answered 4 months ago

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