Exam N10-008 topic 1 question 372 discussion

The network 80.87.78.0/26 represents a block of 64 IP addresses, ranging from 80.87.78.0 to 80.87.78.63. The "/26" indicates that the first 26 bits of the 32-bit IP address are used for the network prefix, leaving 6 bits for host addresses. To determine the range of assignable IP addresses for specific device assignments, we need to identify the valid host addresses within this range. The valid host addresses are 80.87.78.1 to 80.87.78.62, since the network address (80.87.78.0) and broadcast address (80.87.78.63) are reserved and cannot be assigned to a device. Therefore, the answer is C. 80.87.78.1 - 80.87.78.62.

Address - 80.87.78.001010000.01010111.01001110.00000000 Netmask - 255.255.255.192 = 26 = 11111111.11111111.11111111.11000000 Wildcard - 0.0.0.63 = 00000000.00000000.00000000.00111111 Network - 80.87.78.0/26 = 01010000.01010111.01001110.00000000 Broadcast - 80.87.78.63 = 01010000.01010111.01001110.00111111 First IP - 80.87.78.1 = 01010000.01010111.01001110.00000001 Last IP - 80.87.78.62 = 01010000.01010111.01001110.00111110 Hosts/Net - 62 - nnnnnnnn.nnnnnnnn.nnnnnnnn.nnhhhhhh

The answer should be C.

/26 11111111.11111111.11111111.11000000 255.255.255.192 256-192=64 first and last address reserved (64-2=62)

The answer is c

it is C

This answer is C. 80.87.78.1 - 80.87.78.62. If the correct answer is SUPPOSED to be B, how come this is not explained anywhere?? If we get this on the exam, and we choose C, will we get this wrong? very confusing.

C. 80.87.78.1 - 80.87.78.62

It has to be C!

C. 80.87.78.1 - 80.87.78.62 A CIDR notation of /26 refers to the number of bits used for the network prefix. In this case, the network prefix is 26 bits long, which leaves 6 bits for the host addresses. In a CIDR notation /26, the first 26 bits of the IP address represent the network and the last 6 bits represent the host addresses. In this case, the IP address range is 80.87.78.0 - 80.87.78.63 and the network address is 80.87.78.0, so the first usable IP address for devices is 80.87.78.1 and the last usable IP address is 80.87.78.62 A. 80.87.78.0 - 80.87.78.14 is a smaller range than /26 B. 80.87.78.0 - 80.87.78.110 is a bigger range than /26 D. 80.87.78.1 - 80.87.78.158 is a bigger range than /26

The network 80.87.78.0/26 is a Class A network with a subnet mask of /26, which means that it contains 26 bits of network information and 6 bits of host information. The range of valid host addresses for this network is 80.87.78.1 to 80.87.78.62. Any addresses outside of this range are reserved for special purposes or are not used. Therefore, the correct answer is C: 80.87.78.1 - 80.87.78.62. Option A, 80.87.78.0 - 80.87.78.14, includes only a small portion of the valid host addresses for this network, and option B, 80.87.78.0 - 80.87.78.110, includes a much larger range of addresses that goes beyond the valid host addresses for this network. Option D, 80.87.78.1 - 80.87.78.158, also includes a much larger range of addresses that goes beyond the valid host addresses for this network.

Its C, giving you 62 IP's.

Why this answer is correct? This answer is correct because the /26 indicates that there are 26 bits reserved for the network address, leaving 6 bits for the host address. This means that the network address range is 80.87.78.1 - 80.87.78.62, since the first and last address are reserved for the network and broadcast addresses respectively.

C - 62 IP's to use

Answer is C

The answer is C

answer c