From right to left, you have the 1's bit, the 2's bit, the 4's bit and the 8's bit. If there's a 1 in the 1's bit, add 1. If there's a 1 in the 2's bit, add 2. If there's a 1 in the 4's bit, add 4. If there's a 1 in the 8's bit, add 8.
It's 7.
0111
0111 = decimal 7 0011 = decimal 3 ____ 1010 = decimal 10 ____
1. represent every individual digit of given hexadecimal in binary form like this 4---------> 0100 8---------> 1000 7---------> 0111 2. combine the individual binary digits in order to get the binary of given hexadecimal number 487 ------------> 0100 1000 0111 ( required binary number )
1111 1000 ------ 0111 0001 ------ 1000
When you write the decimal number '7' in Base-2 (binary), you write '0111'.
It's 7.
0111
0100 0111 4 7
It is: .0111 that is the largest
1539026015
.0111 is the largest decimal.
the idea of bit stuffing is to replace every two 1's with 110 (redundancy) !! thats because .. to avoid having a 0111 in the middle of the data !! as 0111 is a delimator !! those , we will get (0111) 01101100110110101101101100 (0111) where (0111) is the start and end delimator !!
When the binary number 0111 is incremented by one, it becomes 1000. This is because adding one to 0111 (which represents 7 in decimal) results in 1000, which represents 8 in decimal. The increment causes a carry that changes the last three bits to 0 and adds a 1 to the next left position.
0111 = decimal 7 0011 = decimal 3 ____ 1010 = decimal 10 ____
To represent the decimal number 47 in Binary-Coded Decimal (BCD), you first separate the digits: 4 and 7. In BCD, each digit is represented by its 4-bit binary equivalent. Therefore, 4 is represented as 0100 and 7 as 0111. Combining these, 47 in BCD is expressed as 0100 0111.
.0111%