Examples of intermediate technology
2 answers:
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Answer:
Joe should read the explanatory text and complete the learning activities.
Explanation:
Given
See attachment for options
Required
Best strategy to get through the module
First off, rushing through the activities and taking guess for each question (as suggested by (a)) will not help him;
He may complete the activities but sure, he won't learn from the module.
Also, reading through the units without completing the activities is not an appropriate method because Joe will not be able to test his knowledge at the end of the module.
The best strategy to employ is to read through the units and complete the activities, afterwards (option (b)).
Answer:
Here is the JAVA program that prompts the user to enter positive integers from standard input and prints out the maximum and minimum values:
import java.util.*;
public class Ex0603 {
public static void main(String[] args) { //start of main function
Scanner in = new Scanner(System.in); //creates Scanner class object to take input from user
int minimum = 0; //stores the minimum of the integers
int maximum = 0; //stores the minimum of the integers
int num; //stores the positive integer values
System.out.println("enter a positive integer: ");//prompts user to enter a positive integer
num = in.nextInt();//scans and reads the input integer value from user
minimum = num; //assigns value of num to minimum
while(true) {//while loop keeps prompting user to enter a positive integer
System.out.println("enter a positive integer: ");//prompts user to enter a positive integer
num = in.nextInt();//scans and reads the input integer value from user
if(num <=0)//if the user inputs a negative integer or 0
break; //breaks the loop
if (num > maximum) {//if the value of positive integer is greater than the maximum of value
maximum = num; } //assigns that maximum value of num to maximum variable
if (num < minimum) {//if the value of positive integer is less than the minimum of value
minimum = num; //assigns that minimum value of num to minimum } }
System.out.println("The maximum integer is : " + maximum); //displays the maximum of the positive integers
System.out.println("The minimum integer is : " + minimum); }} //displays the minimum of the positive integers
Explanation:
Here is the JAVA program that that prompts the user to enter an integer N, then reads N double values, and prints their mean (average value) and sample standard deviation:
import java.util.*;
public class Ex0603 {
public static void main(String[] args) { //start of main function
Scanner in = new Scanner(System.in); //creates Scanner class object to take input from user
double integer= 0; //to store the number of input values
double sum = 0; //to store the sum of input numbers
double mean = 0; //to store the average of numbers
double sd = 0; //to store the standard deviation result
double variance = 0; //to store the variance result
double sumSquare = 0; //to store the sum of (num-mean)^2
int n = 0; //to store the sample size N
System.out.println("Enter an integer: "); //prompts user to enter an integer that is how many elements user wants to input
integer = in.nextInt(); //reads the value of integer from user
double size=integer; // declares a double type variable and assigns value of int type variable integer to this double type variable size
int i=0; //used to point each element of the num array
double num[] = new double[(int)size]; //declares an array to hold the numbers input by user
while(integer>0){ //loop executes until value of integer is greater than 0
System.out.println("Enter a number: "); // prompts to enter a number
num[i] = in.nextDouble(); //reads double values
sum += num[i]; //adds the values stored in num array
i++; //increments i to point to the next element in num array
n++; //increments n to count the total number of elements
integer--; //decrements integer value at each iteration
mean = sum / n; } //compute the average of values
i=0;
while(size>0){ //loop executes until value of size exceeds 0
sumSquare += Math.pow(num[i]-mean,2); //takes the sum of square difference between each element of num array and value of mean
i++; //increments i
size--; } //decrements size
variance = sumSquare / (n-1); //compute variance by dividing the result of sum of the squares of their differences from the mean by n-1
sd = Math.sqrt(variance); //computes standard deviation by using sqrt function which takes the sqrt of the result of variance computed above
System.out.println("Average value is: " + mean+ " and the standard deviation is " + String. format("%.2f", sd)); }} //displays the average of numbers stored in mean variable and value of standard deviation is displayed up to 2 decimal places
Screenshots of both programs and their outputs is attached.
