0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
What is the first valid host address of 193.168.22.1?
0001 0110 001 is the first valid host address of 193.168.22.1.
What is valid CLASS-C host address?
An example of a valid class C address would be 192.168.6.200
What is a valid Public IP address for an Internet host?
172.16.15.20
Assuming no subnetting what are valid Class B host addresses?
A classful class B network has a network range of 128 - 191. For host addresses, anything that is legal for an IP address in the last 2 octets would be a valid host address for a class B with no subnets.
What is the valid host range for the subnet in which the IP address 192.168.10.206.255.255.255.240 resides?
The host range is 192.168.10.193 - 206; total usable hosts =14.
Is the ip address 242.1.2.254 a valid address for a host on a tcpip network?
Technically yes, but it's a Class D and most places only use Class A, B or C.
What IP address 10.11.12.13 is a valid address?
The IP address, '10.11.12.13' is not a valid address. This IP address is just a sample that is used.
What is a valid street address for playstation network?
valid address for play station network
What is IP Host address?
The host is the last part of an IP address
How many ip addresses are there in a class A?
There are 2^24 host in class A but (2^24)-2 hosts are valid since the first and last address are reserved .
Why is it unlickely that you will find the IP address 192.168.250.10 on the Internet?
It is unlikely that you will find the IP address 192.168.250.10 on the internet because the .10 represents the number of host connected to an individual network. Also, IEEE recommends that the following IP addresses be used for private networks: 10.0.0.0 through 10.255.255.255 Class A 172.16.0.0 through 172.31.255.255 Class B 192.168.0.0 through 192.168.255.255 Class C
What is the broadcast address for the subnet 10.150.100.96-27?
10.150.100.9627? This isn't a valid IP address ....... BUT, assuming you meant to say 10.150.100.96/27, then the subnetwork address is 10.150.100.96, the 1st host address is 10.150.100.97, and the broadcast address is 10.150.100.127. The /27 means the 1st 27 bits are network, the remaining 5 are host addresses. The subnet address would be 255.255.255.224. If someone could double check my math, it would be appreciated, as I am a newbie too.
