2. When it comes to selling their crop, what are 3 options a farmer has when harvesting their grain?

[10] Create a program called SelectionSort.java that implements the Selection Sort algorithm (The Art of Computer Programming -

Alex787 [66]

Answer:

import java.io.File;

import java.io.FileNotFoundException;

import java.util.ArrayList;

import java.util.Scanner;

public class SelectionSort {

public static void main(String[] args) throws FileNotFoundException {

//For array

ArrayList<Integer>array=new ArrayList<Integer>();

//If argument found

if(args.length>=1) {

//File path

Scanner sc=new Scanner(new File(args[0]));

//Loop until end

while(sc.hasNextLine()) {

//Read each line and add into array

String[] temp=sc.nextLine().split(" ");

for(int i=0;i<temp.length;i++) {

array.add(Integer.parseInt(temp[i]));

}

//Display array

System.out.println("Display array: ");

printArray(array);

System.out.println("\nDisplay array after sort: ");

sortArray(array);

printArray(array);

}

//If argument not found

else {

System.out.println("Argument not found!!!");

}

//Method to print array

public static void printArray(ArrayList<Integer>array) {

for(int i=0;i<array.size();i++) {

System.out.println(array.get(i));

}

//Method to sort array using straight selection sort

public static void sortArray(ArrayList<Integer>array) {

//Step1

for(int j=array.size()-1;j>=1;j--) {

int max=array.get(j);

int index=j;

//Step2

for(int k=j;k>=0;k--) {

if(max<array.get(k)) {

max=array.get(k);

index=k;

}

//Step3

array.set(index,array.get(j));

array.set(j,max);

}

Explanation:

8 0

3 years ago

What are the seven problem solving steps?

Svetradugi [14.3K]

Explanation:

1) Right the problem to solve

2) Anaylse the problem.

3) Define the problem

4) Find solutions

5) Select the best solution

6) Implement the solution

7) Evaluate and learn

3 0

3 years ago

Christopher has designed a fluid power system that repeatedly gets clogs. Which of the following objects should he choose to add

MA_775_DIABLO [31]

Answer:

Valve

Explanation:

Its right

3 0

3 years ago

As part of an insurance company’s training program, participants learn how to conduct an analysis of clients’ insurability. The

sergij07 [2.7K]

Answer:

Part a

The value of cp and cpk for Armand is 0.94 and 0.77 respectively. As both values are less than 1 so Armand is not capable.

The value of cp and cpk for Jerry is 1.42 and 1.33 respectively. As both values are more than 1 so Jerry is capable.

The value of cp and cpk for Melissa is 0.98 and 0.98 respectively. As both values are less than 1 so Melissa is not capable.

Part b

The value of cpk cannot be more than cp. It can at maximum equal to the value of the cp.

Explanation:

Part a

As per the given data

Lower Tolerance Limit=LTL=30 min

Upper Tolerance Limit=UTL=47 min

The process capability ratio is given as

$c_p=\frac{UTL-LTL}{6\sigma}$

where σ is the standard deviation .

The process capability index is given as

$c_p_k=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})$

where μ is the mean.

Armand

Now for Armand, μ is 37.0 , σ is 3.0.

$c_p_{A}=\frac{UTL-LTL}{6\sigma}\\c_p_{A}=\frac{47-30}{6\times 3}\\c_p_{A}=0.94$

$c_p_k_{A}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{A}=min(\frac{47-37}{3\times 3},\frac{37-30}{3\times 3})\\c_p_k_{A}=min(1.1,0.77)\\c_p_k_{A}=0.77$

The value of cp and cpk for Armand is 0.94 and 0.77 respectively. As both values are less than 1 so Armand is not capable.

Jerry

Now for Jerry, μ is 38.0 , σ is 2.0.

$c_p_{J}=\frac{UTL-LTL}{6\sigma}\\c_p_{J}=\frac{47-30}{6\times 2}\\c_p_{J}=1.42$

$c_p_k_{J}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{J}=min(\frac{47-38}{3\times 2},\frac{38-30}{2\times 3})\\c_p_k_{J}=min(1.5,1.33)\\c_p_k_{J}=1.33$

The value of cp and cpk for Jerry is 1.42 and 1.33 respectively. As both values are more than 1 so Jerry is capable.

Melissa

Now for Jerry, μ is 38.5 , σ is 2.9.

$c_p_{M}=\frac{UTL-LTL}{6\sigma}\\c_p_{M}=\frac{47-30}{6\times 2.9}\\c_p_{M}=0.98$

$c_p_k_{M}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{M}=min(\frac{47-38.5}{3\times 2.9},\frac{38.5-30}{3\times 2.9})\\c_p_k_{M}=min(0.977,0.977)\\c_p_k_{M}=0.98$