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Suprisingly, I couldn't find this version of the test anywhere (useful) online. Funny enough, I'd be a first somewhere. The formatting of the question/answers below is simple-- instead of listing all available options, it's just answers.
Question 1
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with 192.168.12.96/27 is:
0.0.0.31
The wildcard mask can be found by subtracting the subnet mask from 255.255.255.255.
Question 2
Fill in the blanks. Use dotted decimal format.
What range of IP addresses is represented by the network and wildcard mask 192.168.70.0 0.0.0.127?
192.168.70.0 to 192.168.70.127
The number of 1s in the wildcard mask represents the number of 0s in the subnet mask. The range of IP addresses for this network would be 192.168.70.0 – 192.168.70.127 with 192.168.70.127 being the broadcast address.'
Question 3
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with 152.115.128.0/17 is
Question 4
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with the network 192.168.12.0/24 is
Suprisingly, I couldn't find this version of the test anywhere (useful) online. Funny enough, I'd be a first somewhere. The formatting of the question/answers below is simple-- instead of listing all available options, it's just answers.
CCNA 2 - Chapter 9.1 QUIZ (100%)
Answers provided by seeseenayy.blogspot.com
Question 1
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with 192.168.12.96/27 is:
0.0.0.31
The wildcard mask can be found by subtracting the subnet mask from 255.255.255.255.
Question 2
Fill in the blanks. Use dotted decimal format.
What range of IP addresses is represented by the network and wildcard mask 192.168.70.0 0.0.0.127?
192.168.70.0 to 192.168.70.127
The number of 1s in the wildcard mask represents the number of 0s in the subnet mask. The range of IP addresses for this network would be 192.168.70.0 – 192.168.70.127 with 192.168.70.127 being the broadcast address.'
Question 3
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with 152.115.128.0/17 is
0.0.127.255
The wildcard mask can be found by subtracting the subnet mask from 255.255.255.255.Question 4
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with the network 192.168.12.0/24 is
0.0.0.255
The wildcard mask can be found by subtracting the subnet mask from 255.255.255.255.
Mask 255.255.255.255
Subnet mask - 255.255.255.0
Wild card mask 0 . 0 . 0. 255
Mask 255.255.255.255
Subnet mask - 255.255.255.0
Wild card mask 0 . 0 . 0. 255
Question 5
Fill in the blanks. Use dotted decimal format.
The wildcard mask that is associated with 128.165.216.0/23 is
0.0.1.255
The wildcard mask can be found by subtracting the subnet mask from 255.255.255.255.
Question 6
Fill in the blanks. Use dotted decimal format.
What range of IP addresses is represented by the network and wildcard mask 172.16.32.0 0.0.15.255?
172.16.32.0 to 172.16.47.255
The wildcard mask 0.0.15.255 would filter addresses smilar to the subnet 255.255.240.0, This wildcard mask would therefore represent addresses from the network address range 172.16.32.0/20 – 172.16.47.255, with 172.16.32.0 being the network address and 172.16.47.255 being the broadcast address.
Question 7
Fill in the blanks. Use dotted decimal format.
What range of IP addresses is represented by the network and wildcard mask 192.168.70.0 0.0.1.255?
192.168.70.0 to 192.168.71.255
The number of 1s in the wildcard mask represents the number of 0s in the subnet mask.
The subnet mask in this case is therefore 255.255.254.0. The range of IP addresses for this network would be 192.168.70.0 – 192.168.71.255 with 192.168.71.255 being the broadcast address.
The subnet mask in this case is therefore 255.255.254.0. The range of IP addresses for this network would be 192.168.70.0 – 192.168.71.255 with 192.168.71.255 being the broadcast address.
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