.0101)2 (b) (16.5)16 (c) (26.24)8 (d) (DABA.B)16
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5) Express the following numbers in decimal:
(a) (10110.0101)2
(b) (16.5)16
(c) (26.24)8
(d) (DABA.B)16
(e) (1011.1001)2
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- Answer the following short questions briefly. Convert the following decimal integers into binary, octal, and hexadecimal:(a) 23, (b) 107, (c) 1238 (d) 92 (e) 173 Convert the following BCD numbers (assume that these are packed numbers) to decimal numbers:(a) 10001001, (b) 00001001(c) 00110010, (d) 00000001 Convert the following decimal numbers into both packed and unpacked BCD forms:(a) 102, (b) 44(c) 301, (d) 1000 Convert the following decimal numbers into 8-bit signed binary numbers:(a) +32, (b) -12 (c) +100 Convert the following binary numbers to the two’s complement form:(a) 1000 0001, (b) 1010 1100(c) 1010 1111, (d) 1000 0000 Find the memory address of the next instruction executed by the microprocessor, when operated in the real mode, for the following CS:IP combinations:(a) CS = 1A00H and IP = B000H(b) CS = 2300H and IP = 1A00H(c)…5. Assume the simple floating point model (i.e., 14-bits of storage) correctlya. How would we represent the numbers 67.25 and 47.98?b. Add the two numbers together. Show the calculation and the final format ofthe result. PS: Please do them in a word processor not hand-written!10. Fill in the following addition table for base 3.+ 0 1 201211. Identify the following statements as TRUE OR FALSE. a) BCD stands for Binary Coded Decimal and encodes each digit of a decimal number to an 8-bit binary form. b) Unicode is a 16-bit code, occupying twice the disk space for text as ASCII or EBCDIC would require. c) A signed-magnitude integer representation includes more negative numbers than it does positive ones. d) A byte is 8 bits, but a word may vary in size (16-bits, 32-bits, etc.) from one architecture to another. e)The largest value that a 60-bit unsigned binary integer can represent is (260-1). f) A 2's complement integer representation includes more negative numbers than it does positive ones.
- MCQS the following? 1) .The number of bytes required to store single precision floating point numbers is a) 2 b) 4 c) 8 d) 16 2) The number of bytes required to store the variable "a", where a = [4 , 6 , -8, 13], would be a) 4 b) 8 c) 16 d) 32 e) 6442. Consider the expression given below. The value of X is:X = 2+9*((3*12)-8)/10 a. 30.0 b. 30.8 c. 28.4 d. 27.2A = {1, 2, 3, 4, 5} B = {1, 3, 5} C = {4, 6} U = {numbers from 0 to 10} 7. 3 ∊ B 8. 5 ∊ C 9. B ⊂ A 10. C ⊂ A 11. C ⊂ B 12. C ⊂ U
- What is the smallest 32-bit floating point number f such that 128 + f > 128 ? What is the smallest 32-bit floating point number g such that 1 + g > 1 ? What is the relationship between f and g?4.Fix the errors.2. The table of ascii characters is of size 128. This requires 7 bits of data. (When stored in an 8 bit byte the high bit is usually left 0) If we had 3 bits of data to represent characters in a data set, how many characters could our set contain? Question 1 options: 7 3 9 8 4
- 6. Convert the following numbers from unsigned binary notation to decimal notation, and from 6-bit 2's complement notation to decimal notation: i) 110011, ii) 001101, iii) 101101 7. Show how each of the following floating point values would be stored using IEEE-754 single precision (be sure to indicate the sign bit, the exponent, and the significand fields): a. 12.5 b. −1.5 c. 0.75 d. 26.625 8. The following is a representation of a decimal floating value using IEEE-754 single precision. Find out the value in decimal. 0 10000011 10101000000...05. Assume the simple floating point model (i.e., 14-bits of storage) correctlya. How would we represent the numbers 67.25 and 47.98?b. Add the two numbers together. Show the calculation and the final format ofthe result. PS: Please do them in a word processor not hand-written nor in a JPEG screenshot Format!!1. Let x be the first two digits of your matric number (group leader), y be the last two digits of your matric number, and define z as x.y If the first digit of y is 0, replace it with 7. Example: Suppose your matric number is 2013104. Thus, x=20, y=74 and z=20.74 determine the IEEE-754 single precision floating-point numbers of decimal value x.y Explain each step.