-2) 215. 168.10.0129 1 Subnet mark 2 No. of Host and Networks 3 first and last...

Transcribed Image Text:

-2) 215. 168.10.0129 1 Subnet mark 2 No. of Host and Networks 3 first and last Hist @Broadcast for each Network 6 QI find the Net ID for the following Ip Address. A-192.168.0.109/28 B- 192.168.0.120/26 c- 200- 168.1. 130/29 Sol # 192.168.0.109. 192.168.0.240 192.168.0.109 | 28 21.1/10000 192.168.0.109 255 255.255 240 Net ID: 192.168.0.96 All Scanned with CamScanner 109/28] Convert to binary El ollollol 11110000 oll 0000096 IP Subnetting class ful: class B for Standard: 172.16.0.0 16 50 classful 1177 CIDR Scanned with CamScanner 18 classless Inter Domin Rout 119 EX for the following IP address find -2) 215. 168.10.0129 1 Subnet mark 2 No. of Host and Networks 3 first and last Hist @Broadcast for each Network 6 QI find the Net ID for the following Ip Address. A-192.168.0.109/28 B- 192.168.0.120/26 c- 200- 168.1. 130/29 Sol # 192.168.0.109. 192.168.0.240 192.168.0.109 | 28 21.1/10000 192.168.0.109 255 255.255 240 Net ID: 192.168.0.96 All Scanned with CamScanner 109/28] Convert to binary El ollollol 11110000 oll 0000096 IP Subnetting class ful: class B for Standard: 172.16.0.0 16 50 classful 1177 CIDR Scanned with CamScanner 18 classless Inter Domin Rout 119 EX for the following IP address find

Answer rating: 100% (QA)

Q 2151681 0029 1subnet mask solution converting to subnet mask The first 29 bits are set to one and the remaining 3 bits are set to 0 in a subnet 29 b... View the full answer