240(10)=11110000(2)
4.1 bit for 2,2 bits for 4,3 bits for 8,4 bits for 16.
This is because the byte values ​​are converted into bits. because you can only work with bits....
In order to know how many bits/second there are in 1 frame/second, you need to know how many bits are in that frame. In a typical asychronous serial protocol with 8 bits per frame, the bit rate would be 0.125 bits/second. If you are talking the IP network layer of TCP/IP, then the frame size is very dependent on the underlying message payload and headers.The original question, by the way, is invalid. Its asks "how many bits does...", but it should have asked "how manys bits per second does...".
for: faster, better readability against: memory isn't used optimally
23 can be represented in binary as 10111 and would therefore require 5 bits to represent.
Eight bits to the octet. The values are 0-255.
16 bits. Java char values (and Java String values) use Unicode.
16 bits per block
16 bits per block
There are 256 possible values (or characters) in 8 bits.
65,536
32 values. 2^5=32
To represent 63 values, you need at least 6 bits, as 2^6 = 64, which can accommodate all 63 values. However, if you're specifically using 8 bits per value, then you would use 8 bits for each of those 63 values, resulting in a total of 63 x 8 = 504 bits.
2^12=4096
To determine how many bits are required to store a specific value, you need to know the range of values that must be represented. The formula to calculate the number of bits (n) needed is ( n = \lceil \log_2(V) \rceil ), where ( V ) is the number of unique values. For example, to store integers from 0 to 255 (256 values), you would need 8 bits, since ( \log_2(256) = 8 ).
8192
1200