Analyze the code below: for i in range(1, 1_): for j in range (1, 2 ): print( 3_, end=_4 ) Output: %23 What should be in the blank in #3?
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[fifth question] Instruction: PROVIDE THE EXACT CODE. BE MINDFUL IN ADDING ADDITIONAL, UNNECESSARY WHITE SPACES.
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- 3: The code on the right is supposed to be the ARM code for procA; however, there are problems with the ARM code. C code: int procA(int x, int y) { int perimeter = (x + y) * 2; I return perimeter; } ARM code: I procA: add rø, r1, r2 mov r1, rø lsl #1 bx lr push {lr} Give the corrected version of the ARM code for procA:int f(int &k){k++;return k * 2;}int main(){int i = 1, j = -1;int a, b, c;a = f(i) + i/2;b = j + f(j) + f(j);c = 2 * f(j);return 0;} What are the values of a, b and c id the operands in the expressions are evaluated from left to right and then what are the values when its evaluated right to left?void main (void) { char result = a; for (char i = { 10; i > 5; i = i-1) result = result + i; } // point B } Consider the code above. What is the decimal value of "result" at "point B" if the variable "a" is initially 59?
- code in java pls Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color's value can range from 0 to 255. Thus (255, 0, 0) is bright red, (130, 0, 130) is a medium purple, (0, 0, 0) is black, (255, 255, 255) is white, and (40, 40, 40) is a dark gray. (130, 50, 130) is a faded purple, due to the (50, 50, 50) gray part. (In other words, equal amounts of red, green, blue yield gray). Given values for red, green, and blue, remove the gray part. Ex: If the input is: 130 50 130 the output is: 80 0 80 Find the smallest value, and then subtract it from all three values, thus removing the gray.Write these in Pseudocode #1a – In pseudocode, write a call to a function that passes 1 Integer variable and 1 Integer array, and saves a Boolean value in return. #1b – In pseudocode, write the function that accepts 1 Integer and 1 Integer array and returns a Boolean. In the function, search the Integer array with a for-loop, and if the Integer parameter is found in the array, return false. If the Integer parameter is not found, return true. #2a – In pseudocode, write a call to a module that passes 1 Integer variable, 1 Real variable, 1 String constant, and 1 String literal as arguments. #2b – In pseudocode, write the module header that accepts 1 Integer, 1 Real, and 2 Strings as parameters. #3 – This pseudocode has multiple problems. Fix the calling statement and the definition below so that the routine accepts 3 grades as parameters and returns the average into a variable.…code this:
- Answer in JavaScript Tim while preparing for his CAT exam, was studying the topic Number Systems. He encountered a loving da question: Given a number N, find the number of pairs (cy) where both x and y are less than N and Highent Common Factor(HCF) of x and y is 1. You have to solve the question for him. Note: 0 can be included within the pairs Input Specification: Input: The number N from which the pairs (x,y) should be obtained Output Specification: Return the total number of pairs Example 1: input1: 4 Output: 9 Explanation: The pairs can be (1,0) ,(0,1) (1,1) (1,2) (2,1) (0, 3) (3,1) (2,3) (0,2) So, a total of 9 pairsCode in python: A certain company has encoded the accounts of its customers and requires that you provide an algorithm that, given an account code, informs if it is valid according to the following description: The account codes are made up of 4 digits counted from right to left, plus the verification digit. The verifying digit is obtained by adding the digits of the account number of the even positions and multiplying the digits of the odd positions, from the new result the residue of the division is extracted for 10, which represents the verifying digit. Develop the code in python as a check digit class that returns 1 if it is correct or 0 if not Heading: Extract digits, verifier calculation, Account verification, Integral algorithmint func(int a, int b) { return (aCSCI 2436:01L Data Structures Lab Lab 1 - Chapter 4 The Efficiency of Algorithms In this lab, you will practice how to measure the running time f a section of code in Java. One approach is to use System.nano Time() where the current time is stored as a long integer equals to the number of nanoseconds. By subtracting the starting time in nanoseconds from the ending time in nanoseconds, you get the run time-in nanoseconds of a section of code. public static void main(String[] args) { int n1 = 10, n2 = 100, n3 = 1000, n4 = 10000; long n1Time, n2Time, n3Time, n4Time; n1Time AlgorithmA (nl); For example, suppose that AlgorithmA is the name of a method you wish to time. The following statements will compute the number of nanoseconds that AlgorithmA requires to execute: } public static long AlgorithmA (int n) { long startTime, endTime, elapsedTime; startTime = System.nanoTime (); int sum = 0; for (int i = 1; i 0 1 2. By midnight, Tuesday, Jan 24th, submit your Java source file and a…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…SEE MORE QUESTIONS