Fill in the following table to show how the given integers are represented, assuming that 16 bits are used to store values and the machine uses two's complement notation. Integ BinHe 4-Byte Big Endian (Hex aryx value as seen in memory) value as seen in memory) 4-Byte Little Endian (Hex er 21 67 5 -1 31
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Q: Perform the following number base conversions. (Tou are required to show your conversion steps…
A: 145.3510=10010001.010110011002 So 145 = 10010001 .35 = .010110011
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A: Answer: Integral part: 01010001 Fractional part: 000001
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A: Find the decimal integer represented by word-length 2’s complement hexadecimal number 6F 20.
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A: SUMMARY: - Hence, we discussed all the points.
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- A(n) __________ is an integer stored in double the normal number of bit positions.Fill in the following table to show how the given integers are represented, assuming that 16 bits are used to store values and the machine uses two's complement notation. Intergers Binary Hex 4-byte Big endian (Hex Value as seen in memory) 4-byte little Endian (Hex value as seen in memory) 28 2216 -18675 -12 31456H - For the IEEE 754 single-precision floating point, write the hexadecimal representation for the following decimal values: (i)–1.0 (ii)– 0.0 (iii)256.015625
- Given a floating point representation 10110 11101101000 (5-bit exponent and 11-bit significant) Question: 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? Please answer based on given numbers.Fill in the following table to show how the given integers are represented, assuming that 16 bits are used to store values and the machine uses two’s complement notation. Integer Binary Hex 4-Byte Big Endian (Hex value as seen in memory) 4-Byte Little Endian (Hex value as seen in memory) 28 2216 −18675 −12 314561. Write down the mathematical notation for fixed point representationand floating representation and also explain each term. 2. Consider the following machine number in 64 bit and Precisely represent the above machine number in decimal digits. 0 10000000011 1011100100010000000000000000000000000000000000000000
- Using the signed-1's complement format, the representation of -7 is_ 1000 0111 1111 1001 This is a standard binary code for the alphanumeric characters that uses seven bits to code 128 characters Hollerith Code ASCII Code EBCDIC Code Gray Code This is defined as a single variable within a term, in complemented or uncomplemented form. coefficient of variable unary variable literal constant1. In a binary coded decimal (BCD) system, 4 bits are used to represent a decimal digit from 0 to 9. For example, 3710 is written as 00110111BCD. (a) Write 28910 in BCD (b) Convert 100101010001BCD to decimal (c) Convert 01101001BCD to binary (d) Explain why BCD might be a useful way to represent numbers3. Fill in the following table to show how the given integers are represented, assuming that 16 bits are used to store values and the machine uses two's complement notation. 4-Byte Big Endian (Hex 4-Byte Little Endian (Hex Integer Binary Hex value as seen in memory) |value as seen in memory) 28 2216 -18675 -12 31456
- Please slove and show all work and steps. Write down the binary representation of the decimal number 63.75, assuming the IEEE 754 double precision format.Assume we are using the simple model for floating-point representation as given in this book (the representation uses a 14-bit format, 5 bits for the exponent with a bias of 15, a normalized mantissa of 8 bits, and a single sign bit for the number). What decimal value (no leading/trailing zeros) for the sum of 01011011001000 and 00111010000000 is the computer actually storing?2. Assume we are using the simple model for floating-point representation as given in this book (the representation uses a 16-bit format, 5 bits for the exponent with a bias of excess-15, a normalized mantissa of 10 bits, and a single sign bit for the number): Show how the computer would represent the numbers a) +3.53125 and -3.34375 using this floating-point format. a) +3.53125 b)-0.09375