What 4-bit value is stored in memory when 110000 utilizing a 1-bit error correcting Hamming ode is received by the memory unit.
Q: In MIPS, all operation codes (op-codes) use ______ only. A- 5 bits B- 8 bits C- 6 bits D -32 bits
A: In MIPS, all operation codes (op-codes) use ______ only. A- 5 bits B- 8 bits C- 6 bits D -32 bits…
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A: 1. if sign is - then take 1 as starting bit for 32-bit representation. 2. 12 in binary is 1100. 3.…
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A: Below is the answer to above question. I hope this will be helpful for you..
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A: Introduction The solution can be achieved as follows.
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A: Actually, 1 byte =8 bits.
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A: Let's consider k be the number of check bits and m be the data bits Given m=720
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A:
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Q: Generator .Determine the out (a) 101010010. (b) 111010011. (c) 110010011.
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A: Here in this question we have asked that Bit rate is the product of ??
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Q: 9-bit binary number
A: Here is the answer which is mentioned below:
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A: Introduction :
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A:
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A: mov AX 3 mov BX 4 mov CX 4
Q: How many check bits are needed if the Hamming error correction code is used to detect single bit…
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A: Actually, binary numbers are nothing but a 0's and 1's.
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A:
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A: According to the question below the solution
Q: Q1: Select any value for AX register b15 b14 b13 b12 bl1 b10 b9 b8 b7 b6 b5 b4 b3 b2 bl b0 X X X X X…
A: As you have posted multiple questions, we will solve the first question for you. Let the AX register…
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A: MAR=4502
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A: Answer : - option : - -128 to 127.
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A: To perform conversion
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A: The answer is given below:-
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A: a) -128 to 127
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A: Required:- On sim8085
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Q: Assuming an 8-bit system is used (i.e. the system uses 8 bits to represent an integer): -6710 -6710
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A: There are 3 questions given.
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A: Let us see the answer below.
Q: A 12-bit Hamming code word containing 8 bits of data and 4 parity bits is read from memory. What was…
A: 1) The 12-bit word is 000011101010 1 2 3 4 5 6 7 8 9 10 11 12 0 0 0 0 1 1 1 0 1 0 1 0 P1 P2…
Q: 3. Complete Twos Complement 8 bit math on the following expression: -43 + 25 4. Complete Twos…
A: Twos complement 8 bit math on the expression -43 + 25 can be evaluated as: Convert -43 into 8 bit…
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- A(n) __________ is a storage location implemented in the CPU.4. Two 8-bit numbers are stored at locations 2000H and 2001H in the memory. Write 8085 code to copy both the numbers at locations 2050H and 2051H respectively. Attach memory view before and after execution. In sim8085 software.Write to and read from the memory device. Write 10 bytes of the information to the memory starting at the address of 0x000008 and read those 10 bytes of information. C code.
- 1.You read the number 11111101111100 from a memory which uses error detection and correction (Hamming algorithm). What was the original correct data word that was written to memory?Create an unsigned 16-bit number that stores the value 13330 (decimal). Show what the memory contains and work verifying the result. Create a signed 16-bit number and store the value -32 (decimal). Show what the memory contains and work verifying the result.The contents of memory location B0000, are FF 16, and those at B0001 6 are 0016. What is the data word stored at the address B0000? Is the word aligned or misaligned?
- The first two bytes of a 4M x 16 main memory have the following hex values: Byte 0: FFByte 1: 01 If these bytes hold a 16-bit two's complement integer, what is the actual decimal value if: a) memory is big endian:__________________ b) memory is little endian:__________________4.6 Find the four hexadecimal values after the statement executes, and the values of NZVC. A: Accumulator (Enter the 4 digit hex content of the accumulator Example FFFF)X: Index Register Enter the 4 digit hex content of the index register Example 00F0)0A3F: Contents of Memory Location 0A3F0A41: Contents of Memory Location 0A41 (Enter the 4 digit hex number Example 1A23)NZVC The status bits (Enter with only 1's and 0's) (a) C90A3F(b) D90A3F(c) F10A41(d) E10A41(e) 790A3F(f) 810A41(g) 990A3F(h) A90A3F(i) 18The first two bytes of a 2M x 16 main memory have the following hex values:Byte 0 is FEByte 1 is 01If these bytes hold a 16-bit two's complement integer, what is its actual decimal value if:a. memory is big endian?b. memory is little endian?
- The first two bytes of a 2M × 16 main memory have the following hex values:1. Byte 0 is FE2. Byte 1 is 01If these bytes hold a 16-bit two’s complement integer, what is its actual decimal value if:1. a) Memory is big endian?2. b) Memory is little endian?Show Steps Please 21. Suppose we want an error-correcting code that will allow all single-bit errors to be correctedfor memory words of length 10. a) How many parity bits are necessary? b) Assuming we are using the Hamming algorithm presented in this chapter to design ourerror-correcting code, find the code word to represent the 10-bit information word:1001100110.What is the alternative term for a single byte?