I would say 3.5 x 1056.
10000. 0000 to 9999
85,000,011 = (8 x 1000000000) + (5 x 100000000) + (0 x 10000000) + (0 x 1000000) + (0 x 100000) + (0 x 10000) + (0 x 1000) + (0 x 100) + (1 x 10) + (1 x 1)
If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as: nPr = n! / (n-r)! 10P4 = 10! / (10-4)! 10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6 10P4 = 7 * 8 * 9 * 10 10P4 = 5040 There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.
9
The given number is the same as 65 and in scientific notation it is 6.5*10^1
149600000 in standard notation is 149,600,000In standard form it is 1.496 × 108
1 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
1 + 1,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111 = 1,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,112 Unless it is binary, in which case: 1 + 111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11111 1111 1111 1111 1111 = 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
0000 0000 0000 0000
its 0000-0000-0000-0000
G'day,It depends upon your word size and notations use.Irrespective of that, a floating point notation has three parts:Sign Bit | Exponent | MantissaSign Bit is always just one bit.Size of Exponent depends upon a particular selection for a notation. Also, the encoding of Exponent could be in different notation, such as 2's comp, biased, excess, etc.Similarly size of Mantissa depends upon the notation.Here, for your question, I will take 32-bit word for FP and use the following convention.Sign Bit | Exponent | Mantissa1 | 8 bits | 23 bits37 in 2's comp is 100101This could be written as 100101.0000____In normalised form, we start populating the Mantissa using first 1 from the leftMantissa: 1001 0100 0000 0000 0000 000If the radix point were at the extreme left of Mantissa, if would have to be moved 6 places to the right to get the original number. So we encode 6 in excess notation using 8 bits. This forms our Exponent.Exponent: 1000 0110Finally, the number is positive, hence the sign bit will be 0, interpreted as per 2's compSign Bit: 0Put this all together.0 1000 0110 1001 0100 0000 0000 0000 00is 37 in floating point.This is correct to the best of my knowledge. If, however, someone has some correction, please do.Cheers,A
0000
The passcode for all Motorola Bluetooth devices is 0000
300 = 256 + 32 + 8 + 4 = Binary 0000 0001 0010 1100
in EBCDIC: 11001000, 10000101, 10010011, 10010011 10010110 in ASCII: 1001000, 1100101, 1101100, 1101100, 1101111 in Unicode: 0000 0000 0100 1000, 0000 0000 0110 0101, 0000 0000 0110 1100, 0000 0000 0110 1100, 0000 0000 0110 1111
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