Question

Java Programming

Question 6 Not yet answered A) What are the parameters of assessment)? Marked out or 20.00 P Flag question Submit the implementation of LAB07 (Project provided to you). Additionally, answer the following questions B) What are the parameters of highestMark) ? C) What is the main condition used to determine the highest mark in the method highestMarko) D) Fill-out the table below using the outcomes of method calls to methodX, see instructions in file provide (Instruction 4) Arguments eg. 5 Outcome 15 E) Submit Word document/PDF or Picture that illustrates the trace to methodX with 5 as an argument

Lab 07 coding

/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package lab07;

/**
*
* @author ckandjimi
*/
public class Lab07 {

/**
* @param args the command line arguments
*/
public static void main(String[] args) {

int outcome = methodX(5);
System.out.printf("Result from Method X= %d",outcome);
String [] studentNames = {"Kingston","John","Nangula","Daniela"};
int Test01[] = {59,72,70,66};
int Test02[] = {80,65,55,49};
  
/**TASKS
* 1. Implement the 2 methods below assessment and highestMark
* 2. Using the 3 arrays declared above invoke or make a method call to assessment
* 3. Invoke highestMark using Test01 or Test02 arrays and store the results in test01Highest,
* test02Highest respectively
* 4. methodX has been fully implemented and Line 19 above makes a call to that function,
* create a table and record the outcomes when the function is called with the following
* arguments : methodX(-5) and methodX(0)
*
*/
  
  
  
}
public static void assessment(/*ADD YOUR PARAMETERS HERE*/){
/**
* This Function takes in 3 arrays one for student names, and 2 for the respective test marks
* We would like to print a list of students names,Test 01,Test 02 and final marks for each
* students(See sample), The final mark is 40% Test 01 and 60% Test 02.
* In the end we would like to print the sub total of all marks and average
* Use the below format for you print out:
* Name Test 01 Test 02 Final
* -------- ------- ------- -------
* Kingston 59 80 72
* John 72 65 68
* ----------------------------------------
* Sub-total 140
* Average 70
* ----------------------------------------
*/
  
}
public static int highestMark(/*ADD YOUR PARAMETERS HERE*/){
/**
* This Functions takes in an array of Marks and return the highest mark
*/
return 0; //Change this statement
}
public static int methodX(int n){
/**
* This function uses recursion to do some interesting calculations
* if the given number(n) is equal or less then 1 the functions returns 1(Stopping case)
* other wise the number(n) is multiplied by the result of the function methodX(n-2)
*/
if (n <= 1 )
return 1;
else
return n * methodX(n-2);
}
}

Answer 1

code self understandable

/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package lab07;

/**
*
* @author ckandjimi
*/
public class Lab07 {

/**
* @param args the command line arguments
*/
public static void main(String[] args) {

int outcome = methodX(5);
System.out.printf("Result from Method X= %d",outcome);
String [] studentNames = {"Kingston","John","Nangula","Daniela"};
int Test01[] = {59,72,70,66};
int Test02[] = {80,65,55,49};
int Test01Highest=highestMark(Test01);
int Test02Highest=highestMark(Test02);
/**TASKS
* 1. Implement the 2 methods below assessment and highestMark
* 2. Using the 3 arrays declared above invoke or make a method call to assessment
* 3. Invoke highestMark using Test01 or Test02 arrays and store the results in test01Highest,
* test02Highest respectively
* 4. methodX has been fully implemented and Line 19 above makes a call to that function,
* create a table and record the outcomes when the function is called with the following
* arguments : methodX(-5) and methodX(0)
*
*/
System.out.println();
assessment(studentNames,Test01,Test02);
System.out.println("Method(-5) = "+methodX(-5));
System.out.println("Method(0) = "+methodX(0));
  
  
}
public static void assessment(String studentNames [],int test01[],int test02[]){
/**
* This Function takes in 3 arrays one for student names, and 2 for the respective test marks
* We would like to print a list of students names,Test 01,Test 02 and final marks for each
* students(See sample), The final mark is 40% Test 01 and 60% Test 02.
* In the end we would like to print the sub total of all marks and average
* Use the below format for you print out:
* Name Test 01 Test 02 Final
* -------- ------- ------- -------
* Kingston 59 80 72
* John 72 65 68
* ----------------------------------------
* Sub-total 140
* Average 70
* ----------------------------------------
*/

System.out.println("Name Test 01 Test 02 Final");
int sub_total=0;
for(int i=0;i<studentNames.length;i++)
{
  
   float x=(float)0.4*test01[i]+(float)0.6*test02[i];
   System.out.println(studentNames[i]+" "+test01[i]+" "+test02[i]+" "+Math.round(x));
   sub_total+=Math.round(x);
}
System.out.println("Sub-total "+sub_total);
System.out.println("Average "+(float)sub_total/studentNames.length);
}
public static int highestMark(int arr[]){
/**
* This Functions takes in an array of Marks and return the highest mark
*/
int highest=arr[0];
for(int i=0;i<arr.length;i++)
{
   if(arr[i]>highest)
   highest=arr[i];
}
return highest; //Change this statement
}
public static int methodX(int n){
/**
* This function uses recursion to do some interesting calculations
* if the given number(n) is equal or less then 1 the functions returns 1(Stopping case)
* other wise the number(n) is multiplied by the result of the function methodX(n-2)
*/
if (n <= 1 )
return 1;
else
return n * methodX(n-2);
}
}

a)The parameters of assesement are String studentNames [],int test01[],int test02[]

b)The parameters of highestMarks is a marks array either Test01 or Test02

c)Make a variable highest equal to first student marks

and for the rest of the students follow this condition if(highest<marks[i]) highest=marks[i]

d)

Arguments outcome

5 15

-5 1

0 1

e)

when n=5

it will return 5*methodX(3)

methodX(3) will return 3*methodX(1)

methodX(1) will return 1

so methodX(3)=3*1=3

and methodX(5)=5*3=15