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In this java program convert to python program then run and show the output. Thank you Source Code: public static void main(String[] args)      {         Scanner in = new Scanner(System.in);         double lowLim ;         double upperLim;         double initAns;         double finAns = 0;                  System.out.println("The function is f(x)= 2x+3(2)");         System.out.println("Width is 0.05\");         System.out.print("Enter Lower Limit: ");         lowLim = in.nextDouble();         System.out.println("");                  System.out.print("Enter Upper Limit: ");         upperLim = in.nextDouble();         System.out.println("");                  double iterations = (upperLim - lowLim)/0.05;                           for(double i=0; i < iterations; i++)         {             System.out.printf("___________________________________________________________________________________________");             System.out.printf("___________________________________________________________________________________________");             System.out.println("");             System.out.println("Iteration: " + (i + 1));             System.out.println("Function \\|\\ Lower Limit \\|\\ Upper Limit \\|\\ LRAM Application (width=0.05) \\|\\ Answer"); //            System.out.printf("f(x)= 2x+3(2) from x=" + "%.0f", lowLim);   //            System.out.printf(" x=" + "%.0f", (lowLim+1));             System.out.printf("f(x)= 2x+3(2) \\|\\ x=" + "%.2f", lowLim);             System.out.printf("\\\|\\ x=" + "%.2f", (lowLim+0.05)); //            System.out.println(""); //            System.out.println("f(x) = (2(" + lowLim + ")+3(2)) * 0.05");             System.out.printf("\\\|\\ f(x) = (2(" + "%.2f", lowLim);             System.out.printf( ")+3(2)) * 0.05");             initAns = ((2 * lowLim) + (3 * 2)) * 0.05;             System.out.printf("\\|\\" + initAns);                          finAns += initAns;             lowLim+=0.05;                          System.out.println("");                      }         System.out.println("");             System.out.printf("___________________________________________________________________________________________");             System.out.printf("___________________________________________________________________________________________");             System.out.println("");          //        System.out.printf("Total Iterations: " + "%.0f", iterations);         System.out.println("");         System.out.println("Final answer/Summation of all iterations is " + finAns);     } } This is sample output below

In this java program convert to python program then run and show the output. Thank you Source Code: public static void main(String[] args)      {         Scanner in = new Scanner(System.in);         double lowLim ;         double upperLim;         double initAns;         double finAns = 0;                  System.out.println("The function is f(x)= 2x+3(2)");         System.out.println("Width is 0.05\");         System.out.print("Enter Lower Limit: ");         lowLim = in.nextDouble();         System.out.println("");                  System.out.print("Enter Upper Limit: ");         upperLim = in.nextDouble();         System.out.println("");                  double iterations = (upperLim - lowLim)/0.05;                           for(double i=0; i < iterations; i++)         {             System.out.printf("___________________________________________________________________________________________");             System.out.printf("___________________________________________________________________________________________");             System.out.println("");             System.out.println("Iteration: " + (i + 1));             System.out.println("Function \\|\\ Lower Limit \\|\\ Upper Limit \\|\\ LRAM Application (width=0.05) \\|\\ Answer"); //            System.out.printf("f(x)= 2x+3(2) from x=" + "%.0f", lowLim);   //            System.out.printf(" x=" + "%.0f", (lowLim+1));             System.out.printf("f(x)= 2x+3(2) \\|\\ x=" + "%.2f", lowLim);             System.out.printf("\\\|\\ x=" + "%.2f", (lowLim+0.05)); //            System.out.println(""); //            System.out.println("f(x) = (2(" + lowLim + ")+3(2)) * 0.05");             System.out.printf("\\\|\\ f(x) = (2(" + "%.2f", lowLim);             System.out.printf( ")+3(2)) * 0.05");             initAns = ((2 * lowLim) + (3 * 2)) * 0.05;             System.out.printf("\\|\\" + initAns);                          finAns += initAns;             lowLim+=0.05;                          System.out.println("");                      }         System.out.println("");             System.out.printf("___________________________________________________________________________________________");             System.out.printf("___________________________________________________________________________________________");             System.out.println("");          //        System.out.printf("Total Iterations: " + "%.0f", iterations);         System.out.println("");         System.out.println("Final answer/Summation of all iterations is " + finAns);     } } This is sample output below

Computer Networking: A Top-Down Approach (7th Edition)
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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In this java program convert to python program then run and show the output. Thank you

Source Code:

public static void main(String[] args) 
    {
        Scanner in = new Scanner(System.in);
        double lowLim ;
        double upperLim;
        double initAns;
        double finAns = 0;
        
        System.out.println("The function is f(x)= 2x+3(2)");
        System.out.println("Width is 0.05\");
        System.out.print("Enter Lower Limit: ");
        lowLim = in.nextDouble();
        System.out.println("");
        
        System.out.print("Enter Upper Limit: ");
        upperLim = in.nextDouble();
        System.out.println("");
        
        double iterations = (upperLim - lowLim)/0.05;
        
        
        for(double i=0; i < iterations; i++)
        {
            System.out.printf("___________________________________________________________________________________________");
            System.out.printf("___________________________________________________________________________________________");
            System.out.println("");
            System.out.println("Iteration: " + (i + 1));
            System.out.println("Function \\|\\ Lower Limit \\|\\ Upper Limit \\|\\ LRAM Application (width=0.05) \\|\\ Answer");
//            System.out.printf("f(x)= 2x+3(2) from x=" + "%.0f", lowLim);  
//            System.out.printf(" x=" + "%.0f", (lowLim+1));
            System.out.printf("f(x)= 2x+3(2) \\|\\ x=" + "%.2f", lowLim);
            System.out.printf("\\\|\\ x=" + "%.2f", (lowLim+0.05));
//            System.out.println("");
//            System.out.println("f(x) = (2(" + lowLim + ")+3(2)) * 0.05");
            System.out.printf("\\\|\\ f(x) = (2(" + "%.2f", lowLim);
            System.out.printf( ")+3(2)) * 0.05");
            initAns = ((2 * lowLim) + (3 * 2)) * 0.05;
            System.out.printf("\\|\\" + initAns);
            
            finAns += initAns;
            lowLim+=0.05;
            
            System.out.println("");
            
        }
        System.out.println("");
            System.out.printf("___________________________________________________________________________________________");
            System.out.printf("___________________________________________________________________________________________");
            System.out.println("");
        
//        System.out.printf("Total Iterations: " + "%.0f", iterations);
        System.out.println("");
        System.out.println("Final answer/Summation of all iterations is " + finAns);
    }
}

This is sample output below

The function is f (x) = 2x+3 (2) Width is 0.05 Enter Lower Limit: 1 Enter Upper Limit: 2 Iteration: 1.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.00 x=1.05 f (x) = (2 (1.0) +3 (2)) * 0.05 0.4 Iteration: 2.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.05 x=1.10 f(x) = (2 (1.05) +3 (2)) * 0.05 0.405 Iteration: 3.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f(x) = 2x+3 (2) x=1.10 x=1.15 E(x) = (2 (1.1)+3 (2)) * 0.05 0.41 Iteration: 4.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f(x) = 2x+3 (2) x=1.15 x=1.20 f(x) = (2 (1.15000000o00000001) +3 (2)) * 0.05 0.41500000000000004 Iteration: 5.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.20 x=1.25 f (x) = (2 (1.2000000000000002) +3 (2)) * 0.05 0.42000000000000004 Iteration: 6.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.25 x=1.30 f(x) (2 (1.250o00000000o0002) +3 (2)) 0.05 0.42500000000000004 Iteration: 7.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.30 x=1.35 f (x) = (2 (1.3000000000000003) +3 (2)) 0.05 0.4300000o00000001 Iteration: 8.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.35 x=1.40 f (x) = (2 (1.3500000000000003) +3 (2)) 0.05 0.43500000000000005 Iteration: 9.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.40 x=1.45 f (x) = (2 (1.4000000000000004) +3 (2)) 0.05 0.44000000000000006 Iteration: 10.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.45 x=1.50 f(x) = (2 (1.4500000000000004) +3 (2)) 0.05 0.44500000000000006 Iteration: 11.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.50 x=1.55 f (x) (2 (1.5000000000000004) +3 (2)) 0.05 0.45 %3D Iteration: 12.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.55 x=1.€0 f(x) = (2 (1.5500000000000005) +3 (2)) * 0.05 0.45500000000000007 Iteration: 13.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.60 x=1.65 f (x) = (2 (1.6000000000000005) +3 (2)) * 0.05 0.4600000000000001 Iteration: 14.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer : f (x) = 2x+3 (2) x=1.65 x=1.70 f (x) = (2 (1.650000000o000006) +3 (2)) * 0.05 0.46500000000000001 Iteration: 15.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.70 x=1.75 f (x) = (2 (1.7000000000000006) +3 (2)) * 0.05 0.47000000000000014 Iteration: 16.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.75 x=1.80 f (x) = (2 (1.7500000000000007) +3 (2)) * 0.05 0.4750000000000001 Iteration: 17.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.80 x=1.85 f (x) = (2 (1.8000000000000007)+3 (2)) * 0.05 0.4800000000000001 Iteration: 18.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.85 x=1.90 f (x) = (2(1.8500000000000008) +3 (2)) * 0.05 0.4850000000000001 Iteration: 19.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.90 x=1.95 f (x) = (2 (1.9000000000000008)+3 (2)) * 0.05 0.49000000000000005 Iteration: 20.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.95 x=2.00 f(x) = (2 (1.9500000000000008) +3 (2)) 0.05 0.4950000000000001 Final answer/Summation of all iterations is 8.95 BUILD SUCCESSFUL (total time: 12 seconds)
Transcribed Image Text:The function is f (x) = 2x+3 (2) Width is 0.05 Enter Lower Limit: 1 Enter Upper Limit: 2 Iteration: 1.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.00 x=1.05 f (x) = (2 (1.0) +3 (2)) * 0.05 0.4 Iteration: 2.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.05 x=1.10 f(x) = (2 (1.05) +3 (2)) * 0.05 0.405 Iteration: 3.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f(x) = 2x+3 (2) x=1.10 x=1.15 E(x) = (2 (1.1)+3 (2)) * 0.05 0.41 Iteration: 4.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f(x) = 2x+3 (2) x=1.15 x=1.20 f(x) = (2 (1.15000000o00000001) +3 (2)) * 0.05 0.41500000000000004 Iteration: 5.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.20 x=1.25 f (x) = (2 (1.2000000000000002) +3 (2)) * 0.05 0.42000000000000004 Iteration: 6.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.25 x=1.30 f(x) (2 (1.250o00000000o0002) +3 (2)) 0.05 0.42500000000000004 Iteration: 7.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.30 x=1.35 f (x) = (2 (1.3000000000000003) +3 (2)) 0.05 0.4300000o00000001 Iteration: 8.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.35 x=1.40 f (x) = (2 (1.3500000000000003) +3 (2)) 0.05 0.43500000000000005 Iteration: 9.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.40 x=1.45 f (x) = (2 (1.4000000000000004) +3 (2)) 0.05 0.44000000000000006 Iteration: 10.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.45 x=1.50 f(x) = (2 (1.4500000000000004) +3 (2)) 0.05 0.44500000000000006 Iteration: 11.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.50 x=1.55 f (x) (2 (1.5000000000000004) +3 (2)) 0.05 0.45 %3D Iteration: 12.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.55 x=1.€0 f(x) = (2 (1.5500000000000005) +3 (2)) * 0.05 0.45500000000000007 Iteration: 13.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.60 x=1.65 f (x) = (2 (1.6000000000000005) +3 (2)) * 0.05 0.4600000000000001 Iteration: 14.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer : f (x) = 2x+3 (2) x=1.65 x=1.70 f (x) = (2 (1.650000000o000006) +3 (2)) * 0.05 0.46500000000000001 Iteration: 15.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.70 x=1.75 f (x) = (2 (1.7000000000000006) +3 (2)) * 0.05 0.47000000000000014 Iteration: 16.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.75 x=1.80 f (x) = (2 (1.7500000000000007) +3 (2)) * 0.05 0.4750000000000001 Iteration: 17.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.80 x=1.85 f (x) = (2 (1.8000000000000007)+3 (2)) * 0.05 0.4800000000000001 Iteration: 18.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.85 x=1.90 f (x) = (2(1.8500000000000008) +3 (2)) * 0.05 0.4850000000000001 Iteration: 19.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.90 x=1.95 f (x) = (2 (1.9000000000000008)+3 (2)) * 0.05 0.49000000000000005 Iteration: 20.0 Function Lower Limit Upper Limit LRAM Application (width=0.05) Answer f (x) = 2x+3 (2) x=1.95 x=2.00 f(x) = (2 (1.9500000000000008) +3 (2)) 0.05 0.4950000000000001 Final answer/Summation of all iterations is 8.95 BUILD SUCCESSFUL (total time: 12 seconds)
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