Convert
0.625 to binary
Translate 0.625 into a fraction. We all know that 0.5 is ½. We know that the remainder, 0.125, is ⅛. Add them together, and you get ½ + ⅛ = ⅝.
Now, in binary, the positions to the right of the point are , which is ½, ¼, and ⅛ respectively.
⅝ is 5 × ⅛. 5 in binary is 101. So, ⅝ is
= 0.101
If you are on a desktop computer you can press the top row key "F3" and a little box will pop up and you can type the word you are looking for in there and it will find that word on the page you are on and take you straight to it.
<span>The statement that P2P networks are most commonly used in home networks is true.
</span>P2P stands for peer-to-peer communication network. It is an example of local administration. <span> Two or more PCs share files and access to devices such as printers without requiring a separate server computer or server software with P2P type of network.</span>
Answer:
The answer is "Option a"
Explanation:
The given question is incomplete, that's why its correct solution can be defined as follows:
The above-given question is the part of the Binary Autocomplete, in which this Autocomplete function would be a full word or sentence after just a few other letters were entered into the system. It approach improves text taking appropriate on smartphone devices of particular because every letter should not be written in such a single phrase.
Answer:
1. The network address for 172.22.49.252/17 is 172.22.0.0/17.
2. The last valid assignable host address of 172.22.4.129/26 is 172.22.4.190.
3. The first and last host address of 192.167.25.25/16 is 192.167.0.1 and 192.167.255.254.
4. The broadcast address of 10.75.96.0/20 is 10.75.111.255
Explanation:
Subnetting in networking is the process of managing the use of host addresses and subnet masks of a network IP address. For example, the IP address "172.22.49.252/17" is a class B address that receives an extra bit from the third octet which changes its subnet-mask from "255.255.0.0" to "255.255.128.0". with this, only 32766 IP addresses are used, with the network address of "172.22.0.0/17".
