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- A(n) __________ is a storage location implemented in the CPU.Most Intel CPUs use the __________, in which each memory address is represented by two integers.The 32-bit number 52AB43FC (in hexadecimal) is stored in abyte-addressable memory starting at physical address FE08 (in hex) using Little-Endian notation. The byte(value ni hexadecimal) stored at theaddress FE0B will be
- A computer has a 16 bit address space A[15:0]. If all addresses having bits A[15:14]=11 are reserved for I/O device registers, then the maximum number of actual word addressable memory locations is 216 - 214 Group of answer choices True Falsemcq question The Physical address of the String Destination Memory Reference is: DS:DI, DS:SI, and DS:BX SS:SP, and SS:BP ES:DI only ES:DI and ES:SIAssembly language programming in MIPS. Use QTSpim to run code. Write a simple Assembly Language program that has a data section declared as follows:.data.byte 12.byte 97.byte 133.byte 82.byte 236 Add the values up, compute the average, and store the result in a memory location.
- Micro processor 8086 Assume that there are two bit 32-bit numbers:4444DDDDH and CCCC0003H .write an assembly program program to sum them , if it’s possible to do it with intel 8086 microprocessor then cross check (4444DDDH and CCCC0003HIn sim8085. c) Execute the given 8085 instructions and attach the memory view before and after execution. Take value of your choice at the required memory locations. Code: LXI H, 2000H MOV A,M ADI 02H INX H MOV M,A HLTthe available space list of a computer memory is specified as follows: 9 start address block address in words 100 50 200 150 450 600 1200 400 determine the available space list after allocating the space for the stream of requests consisting of the following block sizes: 25,100,250,200,100,150 use i) first fit ii) best fit and iii) worst fit algorithms
- In a 1 MB memory divided into 64 KB segments, if a segment starts at the address 1234A find the last address in the segment.A memory that is capable of determining whether a given datum is contained in one of its addresses is CAM why?0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.