The most important result in all of statistics is the .  The answer is three words and is covered in Module 5.  

Assume the operating system for our simulator has had new service routines added to it. Consider the following code fragment that requests one of those new service routines: LW a2, %lo(B)(t0) ADDI a0, zero, 34 LOOP: ECALL ADDI a2, a2, -1 Also consider the following partial trap vector table (aka jump table):    ADDRESS      CONTENTS   0x0000_007C 0x0000_0708 0x0000_0080 0x0000_0528 0x0000_0084 0x0000_0914 0x0000_0088 0x0000_0958 A. What address is used by the code above to access the trap vector table?  B. What is the starting address of the service routine called by the code above?  C. If the symbol table shows LOOP is address 0x0003_5850, what is the return address from the service routine?  D. What is the machine code for the instruction the service routine uses to return to the code above?  Enter the addresses and the machine code as 8-digit hex values in RV32I format like the following: 0x0000_01F3

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes: ADDRESS CONTENTS … … 0x0C10 0101 0010 0111 1011 0x0C0F 0000 0000 1010 0101 0x0C0E 1010 0101 0000 0011 0x0C0D 0011 1110 1001 0100 0x0C0C 0010 0011 0110 0010 0x0C0B 0000 1100 0000 1101 … … The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) Interpret the contents at address 0x0C0C as two ASCII characters. B.) Interpret the contents at the same address as A.) above as an unsigned integer in base 10. C.) Assume address 0x0C0F holds the more significant digits of a value (the high 16 bits) and address 0x0C0E holds the least significant digits of the same value (the low 16 bits). Interpret the resultant value an RV32I instruction to determine the operation.(Enter either ADD, JR, LW, or OTHER if it is not one of the first 3) Recall that a memory location can store an address. We call that memory location’s contents a “pointer” since it’s an address that “points” to another memory location. D.) Interpret the contents at address 0x0C0B as a pointer.(Enter hex like the following example: 0x2A3F) E.) What are the contents of the memory location that the pointer above is pointing to?(Enter hex like the following example: 0x2A3F)   REFERENCE: Partial Table of Hex to ASCII Characters: 20 sp 30 0 40 @ 50 P 60 ` 70 p 21 ! 31 1 41 A 51 Q 61 a 71 q 22 ” 32 2 42 B 52 R 62 b 72 r 23 # 33 3 43 C 53 S 63 c 73 s 24 $ 34 4 44 D 54 T 64 d 74 t Table of RV32I Opcodes in Hex: ADD 0x33 JR 0x67 LW 0x03

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes: ADDRESS CONTENTS … … 0x0C10 0101 0010 0111 1011 0x0C0F 1111 1111 1111 0101 0x0C0E 0111 0101 0001 0011 0x0C0D 0110 1010 0001 1111 0x0C0C 0111 0000 0101 0100 0x0C0B 0000 1100 0000 1101 … … The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) Interpret the contents at address 0x0C0C as two ASCII characters. B.) Interpret the contents at the same address as A.) above as an unsigned integer in base 10. C.) Assume address 0x0C0F holds the more significant digits of a value (the high 16 bits) and address 0x0C0E holds the least significant digits of the same value (the low 16 bits). Interpret the resultant value an RV32I instruction to determine the operation.(Enter either ADD, JR, LW, or OTHER if it is not one of the first 3) Recall that a memory location can store an address. We call that memory location’s contents a “pointer” since it’s an address that “points” to another memory location. D.) Interpret the contents at address 0x0C0B as a pointer.(Enter hex like the following example: 0x2A3F) E.) What are the contents of the memory location that the pointer above is pointing to?(Enter hex like the following example: 0x2A3F)   REFERENCE: Partial Table of Hex to ASCII Characters: 20 sp 30 0 40 @ 50 P 60 ` 70 p 21 ! 31 1 41 A 51 Q 61 a 71 q 22 ” 32 2 42 B 52 R 62 b 72 r 23 # 33 3 43 C 53 S 63 c 73 s 24 $ 34 4 44 D 54 T 64 d 74 t Table of RV32I Opcodes in Hex: ADD 0x33 JR 0x67 LW 0x03

Complete the following program that only adds integers that are odd positives. The program considers the NON-ZERO integers at labels N1, N2, and N3 and stores the sum of the appropriate integers in the memory at label SUM. It uses a subroutine at label CHKN to check each integer. .origin 0x0001_0000 START: LUI t0, %hi(START) LW a0, %lo(N1)(t0) L1: ADDI a1, a0, 0 LW a0, %lo(N2)(t0) ADDI a6, t0, %lo(CHKN) JR a6 L2: ADD a1, a1, a0 LW a0, %lo(N3)(t0) L3: ADD a1, a1, a0 SW a1, %lo(SUM)(t0) HALT CHKN: ADDI a4, a0, 0 # move in to a4 for testing ANDI a4, a4, 0x0001 # right most bit is always 1 if odd WRONG: ANDI a0, a0, 0 # zero out wrong values RETURN: RET N1: .word # value not shown N2: .word # value not shown N3: .word # value not shown SUM: .word 0

Consider the following IEEE 754 32-bit floating point value: 10010011_00011000_00000000_00000000 Enter S, its sign, as either + or -. Enter E, its exponent part’s value, in base 10 (not including the -127 bias). Enter F, its fraction part’s value, in base 10 in the form 0.XXXX where X is a digit.

The transverse plane divides the body into superior and inferior halves.

The skin is deep to the bones of the body.

Testing your computer through a browser is:

A tape measure can be used to measure AROM for all of the following, EXCEPT which?