Respuesta :
Answer:
255.255.255.0
Explanation:
Given data
base IP address = 198.10.0.0/20
receives addressed = 198.10.10.144
to find out
network mask used by the router
solution
we know mask is combination of 32 bit used
describe to portion in addressed refers to subnet that is
we know IP address = 198.10.0.0/20
here a = 198, b = 10 and c = 0 and d = 0 and n = 20
so it will be 20 bits + 12 bits
20 bits are 1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1. 1 1 1 1 1 1 1 1
12 bits are 0000.0000.0000
so default natural mask are
Class A is 255.0.0.0
Class B is 255.255.0.0
Class C is 255.255.255.0
Answer:
255.255.240.0
Explanation:
The base IP address is 198.10.0.0/20 and the Subnet Masks is given (/20), we can transform this subnet mask to the standard notation by converting the /20 to binary form as follow -> A subnet mask is made of 32 bits, /20 indicates that the first 20 bits are one, 11111111 11111111 11110000 00000000 converting to decimal 255.255.240.0.
Our base IP is 198.10.0.0
Our netmask is 255.255.240.0
Our HostMin is 198.10.0.1
Our HostMax is 198.10.15.254
Therefore, the received packet, 198.10.10.144, is in the range of IPs