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- Create a C# application that generates a BitArray of five Byte values (1,2,3,4,5) and displays each byte in its correct binary form (attach output snapshot).5. Given the descriptor: Determine the following: a. Limit: b. Starting Address: c. Ending Address: d. Segment: e. DPL: f. G: g. AV: h. D: i. Descriptor used: 7 00H 5CH 6 5 9AH 01H 4 5678H 1 1234HWhat will be the value stored in address 0007H when the given code is executed. Given value at 0002H= 8, 0003H = 4 LDA 0002H MOV B,A LDA 0003H ADD B STA 0007H hlt
- adres veri 01h 5x9 02h 5x8 8715683 b190100564 7156 08h 6715988 5x2 bis010054 71 09h 5x1 Write the asm code that will create the address and contents given in the table. (It is mandatory to use loop and indirect addressing.) b10100564 159 190100564-7I5A CPU that supports little endian format reads two integer (4-byte) values from address 0x1000 and 0x2000. The values read are 55 and 6850 respectively. Please show the memory contents (byte-wise) at address 0x1000 and 0x2000?In the event that each page is 1 KB in size, what are the page numbers and offsets (please supply them as decimal integers) for the following address references? (please provide the proper page numbers and offsets, as well as the stages of the computation)a. 2562 b. 23563 c. 11500
- How many block pointers will be required for a text file with 100,000 ascii characters with a block size of 4096 bytes and a block pointer size of 64 bits?Write a program to add 32-bit number in memory locations staring at [0000] to 32-bit number in memory locations starting at [0008], and put the result in memory locations starting at [0010].LCPU assignment Multiply the number by 1.25: A = X * 1.25X = 0x3C (direct). Then save the result to main memory using direct addressing. Example: Multiply the value by 0.75: X * 0.75 = 0.5 * X + 0.25 * X X = 100 (0x64)100 * 0.75 = 750x64 = 01100100 00110010 [Shift 1x to the right - add 0 to the left and cut off one bit from the right]00011001 [Shift 2x to the right - add 00 to the left and truncate 2 bits from the right]01001011 [Sum of previous two transactions] 01001011 = 0x4b = 75
- MIPS Assembly: Below is the code and attached is the syntax error it has. Please fix the syntax error and print the output. Attached is image of what the output must be. Data.asm .dataarr: .word 7, 9, 4, 3, 8, 1, 6, 2, 5n: .word 9 n .text.globl mainmain:la $t0, arr # Load address of arrla $t1, n # Load address of nlw $t2, 0($t1) # Load n into $t2 jal radixSort # Call radixSort la $a0, arr # Load address of arrlw $a1, 0($t1) # Load n into $a1jal printData # Call printData li $v0, 10 # Exit the programsyscallradixSort:la $a0, arr # Load address of arrlw $a1, 0($t1) # Load n into $a1 jal getMax # Call getMax move $t3, $v0 # Move the max to $t3move $t4, $zero # Move 0 to $t4 loop:ble $t3, $zero, Exit # Check if m/exp <= 0 mul $t5, $t4, 10 # Multiply exp by 10jal countSort # Call countSort addi $t4, $t4, 1 # Increment expdiv $t3, $t3, 10 # Divide m by 10j loop # Loop back Exit:jr $ra # ReturngetMax:move $t6, $zero # Move 0 to…The following data segment starts at memory address 0x4100 (hexadecimal).data printString BYTE "Do not add decimal to hex",0 someBytes WORD 36 DUP(0) moreBytes BYTE 10, 20, 30, 40, 50, 60, 70, 80, 90 questionAddr DWORD ? ignoreMe WORD? Ans:-Assume the range of data in a sequence of data stream is148,000,000,050 to 1.75. What is the minimum number of binary bitsyou need to represent the data range.