Using 32-bit IEEE 754 single precision floating point with one(1) sign bit, eight (8) exponent bits and twenty three (23) mantissa bits, show the representation of -11/16 (-0.6875).
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- Given two 32-bit single-precision Floating-point number representation (IEEE-754) in normalized form, Give their sum (A+B) in IEEE-754. A: 11000000010110100101010000000000 B: 11000000010110100101010000000000Using 32-bit IEEE 754 single precision floating point with one(1) sign bit, eight (8) exponent bits and twenty three (23) mantissa bits, show the representation of -11/16 (-0.6875).Given the following in 32-bit single-precision Floating-point number representation (IEEE-754) in normalized form, A:11000000010110100101010000000000 B:01000001100111001000101010000000 Find the sum of A and B (A+B) in IEEE-754 form
- Given a floating point representation 10110 11101101000 (5-bit exponent and 11-bit significant) 3. if the exponent is in signed magnitude and the significant is in two’s, what are its normalized floating point representation in hexadecimal and its real value in hexadecimal?Given a floating point representation 10110 11101101000 (5-bit exponent and 11-bit significant) if the exponent is in two’s and the significant is in one’s, what are its normalized floating point representation in hexadecimal and its real value in hexadecimal?Using 8-bit representation, what is the 2’s complement of the result of ((45 base 10) + (44 base 10))?
- Express -1/32 in the IEEE 754 single precision format.Perform the arithmetic operations (36)10+(-41)10 and (-1F)16(-2D)16 in .A binary using 12-bit signed-2's complement representation for negative numbers. Convert the answers back to decimal and verify that they are correct Convert the decimal number to single precision IEEE 754 floating-point .B .representation. Show all your work8. In IEEE Single precision, for which values of the exponent e, is r = 2e: A Normalised element e Fs(2, 24, -126, +127) A Denormalised element