Pls answer only if u know, correct answers only pls! will rate you if correct 1.) What is the address of the last (bottom-rightmost) memory location? My answer is 4095 but it is wrong. I thought MARIE starts at 0000 and ends with 4095 so I need help on this. 2.) What is the content of memory address 001? (No need to write 0x from hereon; we already assume hexadecimal unless otherwise stated.) My answer is 3000 but it is wrong

Suppose a program’s 15-th logical instruction (counting starts from zeroth) is at physical address 1234ABDE16 in the RAM. (contents of any logical address fits into any physical address) What is the physical address in the RAM where the program has been loaded?  What is the physical address in the RAM of the last instruction of the program if it has 37 logical instructions altogether?

I have this problem from my textbook that I do not understand, despite re-reading the section on segmentation and paging. "The IBM system/370 architecture uses a two-level memory structure and refers to the two levels as segments and pages, although the segmentation approach lacks many of the features described in Chapter 8. For the basic 370 architecture, the page size may be either 2 KB or 4 KB, and the segment size is fixed at either 64 KB or 1 MB. For the 370/XA and 370/ESA architectures, the page size is 4 KB and the segment size is 1 MB. What advantages of segmentation does this scheme lack? What is the benefit of segmentation for the 370?"Can you help me undertand what they are looking for in this explanation?

Indirect Addressing Mode Instruction 002A J@ RETADR 3E20030030 RETADR RESW 1 1.Instruction Starts with 002A2.Opcode of J is 3C3.Object code is 3E2003Please show the details

Explain me please step by step the answer is pc is 007 and ac 0077

Given a computer with 300 memory locations, each 18-bit assessable, how many bits of addressing are needed?  Given a computer with 10-bit addressability and 5 bits for the address, what's the maximum number of memory this computer can have in bits?

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mcq 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:SI

Address Instruction Ox242 CALL sort Ох262 sort: MOVLW Ох90 Fill in the following blanks based on the above code segment: First word of machine code stored in locations 0x243 and Ox242: Ox * Second word of machine code stored in locations Ox245 and Ox244: Ox