Answer:
iooooooooooooooooooooo
Explanation:
ooooooooooooooo
oooo9ooooo9oooooooooo
ooooooooooooooppppooooooooooo
oooooooooooo
ooooooooooooooooooooo
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
Answer: False
Explanation:
The statement that "Peacekeeper text is non-printing text that indicates where you can type" is false.
The non-printing text which shows where on exam type is referred to as the peaceholder text. It is usually an hint which can be used to fill in the actual text.
Answer:
B I hope I'm right sorry if I'm wrong
