I i. In. *M & 4 9:YQ1/(7 Marks) /For the circuit shown below,:(select the write answer)1.2 Ωww20 A80 12062(2) ibäi The total resistance (equivalent) can:be found as[{[(12//8)+1.2]+[4+(6//3)]}+1]//2 OK[(12//8)+1.2]//[4+(6//3)]}+1]//2 OK[(12//8)+(1.2//4)+(6//3)]+1}//2] OK[[(12//8)+(1.2+4)+(6//3)]//1}//2] Oنقطة واحدة:The total resistance value is2.402.3 Q.1.3 Q1.5 QII<>wwwwww

It is defined as The electrical resistance of a circuit component or device is defined as the ratio…

Answered: (aua resistance (equivalent) can :be…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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write in assebly language using lc3 simulator LC3 simulator-https://wchargin.com/lc3web/ Write a program that implements a for loop that adds 5 to R3 every iteration a. Increment a register by 1 in every cycle of the loop to keep track of the loop iterationsb. Store the maximum amount of iterations that the loop can have in a memory locationc. Your for loop should start at 0 iterations (i.e. i = 0) and stop looping once the maximum is reached Psuedo code: .ORIG x3000 ; Step 1: Clear R3 ; Step 2: Clear R0 ; Step 3: Load the value of MAX into R1 (maximum amount of iterations) ; Step 4: Take the two's compliment of R1 (2 instructions!) loop ; Step 5: Add #5 to R3 (result) ; Step 6: Increment R0 (loop iteration counter) ; Step 7: Add R1 and R0 and put the result in R2 (Since we did a 2's complement on R1, we are essentially subtracting R1 from R0) ; Step 8: Branch if negative back to loop (this means that R1 > R0 and we need to keep looping) ; Step 9: Halt CPU…
Consider the code presented below: y = (1, 9, 2) y[1]=8 print(y) What do we get when we execute this code? Answer:
Consider the following context-free grammar for statements: 1) E > E+T 2) E →T 3) T > I*F 4) T →F 5) F → (E) 6) F → id which can be analyzed with the LR Parsing Table given as following. Action Goto id + * () $ ETF State s5 s4 |12 3 1 s6 асс 2 r2 s7 r2 r2 3 r4 r4 r4 r4 4 $5 $4 8 2 3 5 r6 r6 r6 r6 6. s5 $4 9 3 7 s5 $4 10 s6 s11 9 rl s7 rl rl 10 r3 r3 r3 r3 11 r5 r5 r5 r5 Show a complete parse, including the parse stack contents, input string, and action for the string id*(id + id), using the grammar and parse table given. 8.
Example function void f() { int array[10]; read(array, 10); reverse(array, 10); print(array, 10); } Stack Frame saved $ra = 4 bytes int array[10] (40 bytes) Figure 7.6: Example of a function f and its corresponding stack frame 5. Suppose we rewrite function f (Figures 7.6) to have an integer parameter n. The local array is now declared to have n integers (rather than 10). This means that the size of the stack frame size of function f will depend on n. Rewrite the function f in MIPS assembly language. Hint: you may use the $fp register (in addition to $sp) to implement the function f. void f(int n) { int array[n]; read(array, n); reverse(array, n); print(array, n); }
(10) Solve the following recurrence (you may assume n is power of 4). Use backward substitution: T(n) = 2T ( T (17) + + 3n кг; чи аромисов T(1) = 1 су , не
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#include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…
import java.lang.System; 3. public class FibonacciComparison { 4. // Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8 .... /* 7 input cases 8. 1) 0 9 2) 3 10 3) -1 11 4) 9 12 output cases 13 1) 0 14 2) 2 15 3) 0 16 4) 34 17 */ // Note that you need to return 0 if the input is negative. // Please pay close attention to the fact that the first index in our fib sequence is 0. 18 19 20 // Recursive Fibonacci public static int fib(int n) { // Code this func. 21 22 23 24 return -1; 25 26 // Iterative Fibonacci 27 28 public static int fiblinear(int n) { // Code this func. 29 30 return -1; 31 32 33 public static void main(String[] args) { 34 // list of fibonacci sequence numbers int[] nlist w { 5,10, 15, 20, 25, 30, 35, 40, 45}; 35 36 37 // Two arrays (one for fibLinear, other for fibRecursive) to store time for each run. // There are a total of nlist.length inputs that we will test double[] timingsEF = new double[nlist.Length]; double[] timingsLF = new double[nlist.length]; 38 39 40 41 42 // Every…
java
Problem overview As you may already know from earlier courses, integer values in C++ (as well as other programming languages) are limited by the number of bits used to represent these data. For instance, a 64-bit unsigned integer has the maximum value of 2^64 - 1 or 18446744073709551615. One method of representing integers of arbitrary length, which was covered last spring, is a a linked list data structure such that that a node in the linked list corresponds to a single digit in the integer. For instance, the number 123 can be stored as linked-list that could look like this: [ 3 ] -> [ 2 ] -> [ 1 ] -> NULLNote that the first (or head) node in this list contains the least significant digit (in this case 3), while the last node contains the most significant digit (1). For this problem, you are to read in pairs of arbitrary length integers and produce correct output as outlined below. InspirationNote, this problem is inspired by Problem 8.19 from Elements of Programming…
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(aua resistance (equivalent) can :be found as The tota {[(12//8)+1.2]+[4+(6//3)]}+1]//2 O E{(12//8)+1.2]/[4+(6//3)]}+1]//2 O K[(12//8)+(1.2//4)+(6//3)]+1}//2] O KI(12//8)+(1.2+4)+(6//3)1//1}//21