Imagine a decimal number, for example, 123 (hundred and twenty-three). Each digit has a corresponding place-value; the right-most digit has the place-value 1, the next digit (counting from the right) has the place-value 10, the next digit hast eh place-value 100. The right-most position (where the digit "3" is in this example) is in the position of least value - the least significant position.
When several bits represent an integer, the situation is the same, except that the numbers are in base-2 instead of base-10 (each position is worth twice as much as the position to the right). But you still have the concept of place-value, and the digit that represents the 1's position is the "least significant bit".
multiplication/division: least number of significant figures addition/subtraction: least number of numbers to the right of decimal point
The final answer.is only as accurate as the least accurate component in the calculation, so use the significant figures of the measurement with the fewest.
yes at least a little bit
12.5912
When adding or subtracting significant figures(sig figs), the answer will be significant to the same number of decimal places as the number with the least number of decimal places used in the calculation. Example: 12.44+1.6+133.887=147.927 ==>147.9
Least Significant Bit. Least Significant Byte. (Depends on use.)
1
I assume you mean a binary representation of a number.The "least significant bit" (usually the one to the far right but in some languages it has another placement) is "ones"the next most significant bit are the twosThe third most significant bit are the foursetc.So if your number is 37there is one 32 (the sixth most significant bit)no 16's (the fifth most significant bit)no 8's (the fourth most significant bit)one 4 (the third most significant bit)no 2's (the second most significant bit)one 1 (the least most significant bit)if we are to fill an 8 bit "word " we get:0010 0101
Least significant bit.
It is not. It is probably the least significant question that I have answered! It is not. It is probably the least significant question that I have answered! It is not. It is probably the least significant question that I have answered! It is not. It is probably the least significant question that I have answered!
32 bit
The Bit/Byte order (endianess) of a computer is dependent on the central processing unit (CPU). The large portion of personal computers today are based on the x86 architecture and use a little-endian byte order and least significant bit (LSB) bit order.
The bits in a numeric value like 00000000 00110011 have a decimal value based on the bit position. The most significant bit is the one that has highest decimal value and is the left most bit. The right most bit is the least significant bit. High-order bits are the half of the number of bits that have the highest values, the left most bits in the 16 bit value above The low order bits in this case are the right most bits. This should not be confused with bit placement in memory/cpu registers. Intel/AMD cpus are little edian, meaning that the most significant part is physically right and the lest significant is left most (the bits are not in reverse order). Google for a more detailed info.
All factors are equally significant.
bool is_even(long int num) { return !(num & 1); //when the number is even(divisible by two), //its least significant bit is 0 }
0111-1111
1000000) has 1 significant figure and a least significant decimal of 6.