Write a RainFall class that stores the total rainfall for each of 12 months into an array of doubles. The program should have methods that return the following:
the total rainfall for the year
the average monthly rainfall
the month with the most rain
the month with the least rain
Demonstrate the class in a complete program.
Input Validation: Do not accept negative numbers for monthly rainfall figures.
import java.io.*;
import java.util.*;
public class Rainfall
{
Scanner in = new Scanner(System.in);
private int month = 12;
private double total = 0;
private double average;
private double standard_deviation;
private double largest;
private double smallest;
private double months[];
public Rainfall()
{
months = new double[12];
}
public void setMonths()
{
for(int n=1; n <= month; n++)
{
System.out.println("Enter the rainfall (in inches) for month #" + n + ":" );
months[n-1] = in.nextInt();
}
}
public double getTotal()
{
total = 0;
for(int i = 0; i < 12; i++)
{
total = total + months[i];
}
System.out.println("The total rainfall for the year is" + total);
return total;
}
public double getAverage()
{
average = total/12;
System.out.println("The average monthly rainfall is" + average);
return average;
}
public double getLargest()
{
double largest = 0;
int largeind = 0;
for(int i = 0; i < 12; i++)
{
if (months[i] > largest)
{
largest = months[i];
largeind = i;
}
}
System.out.println("The largest amout of rainfall was" + largest +
"inches in month" + (largeind + 1));
return largest;
}
public double getSmallest()
{
double smallest = Double.MAX_VALUE;
int smallind = 0;
for(int n = 0; n < month; n++)
{
if (months[n] < smallest)
{
smallest = months[n];
smallind = n;
}
}
System.out.println("The smallest amout of rainfall was" + smallest +
"inches in month " + (smallind + 1));
return smallest;
}
public static void main(String[] args)
{
Rainfall r = new Rainfall();
r.setMonths();
System.out.println("Total" + r.getTotal());
System.out.println("Smallest" + r.getSmallest());
System.out.println("Largest" + r.getLargest());
System.out.println("Average" + r.getAverage());
}
}

Automation tools allow applications, businesses, teams or organizations to automate their processes which could be deployment, execution, testing, validation and so on. Automation tools help increase the speed at which processes are being handled with the main aim of reducing human intervention.

Linkedln automation tools are designed to help automate certain processes in Linkedln such as sending broadcast messages, connection requests, page following and other processes with less or no human or manual efforts. Some of these automation tools include;

i. Sales navigator for finding right prospects thereby helping to build and establish trusting relationships with these prospects.

ii. Crystal for providing insights and information about a specified Linkedln profile.

iii. Dripify used by managers for quick onboarding of new team members, assignment of roles and rights and even management of subscription plans.  

Answer:

Well, for me I personally use LinkedCamap to drive more LinkedIn connections, hundreds of leads, sales, and conversions automatically.  

Some other LinkedIn automation tools are;  

Hope this helps!

Answer:

c. Bond Spread

Explanation:

An interactive chart is used to show all the details in the chart and the user can extend or shrink the details that is presented in the charts by using the slider control.

The bond spread in an interactive is a chart that is used to compare and chart the current spread between the corporate bond as well as the benchmark government bond.

In Interactive Charting, the chart type that give room for charting the current spread between a corporate bond and a benchmark government bond is C:Bond Spread.

An interactive charts serves as a chart that give room to the user to carry out zooming actions as well as hovering a marker to get a tooltip and avenue to choose variable.

One of the notable type of this chart is Bond Spread, with this chart type, it is possible to charting the current spread that exist between a corporate bond as well as benchmark government bond.

Therefore, option C is correct because it allows charting of the current spread between a corporate bond.

Learn note about interactive charts at:

https://brainly.com/question/7040405

Answer:

Well, for me I personally use LinkedCamap to drive more LinkedIn connections, hundreds of leads, sales, and conversions automatically.

Some other LinkedIn automation tools are;

Hope this helps!

Answer:

supercomputers are the computer that is used in scientific research.

Answer:

super computer is the right answer

Answer:

removeDuplicates() function:-

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

        //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

Explanation:

Complete Code:-

//Create Linked List Class.

class LinkedList {

       //Create head pointer.

       static ListNode head;

       //define structure of ListNode.

       //it has int val(data) and pointer to ListNode i.e, next.

   static class ListNode {

       int val;

       ListNode next;

       //constructor to  create and initialize a node.

       ListNode(int d) {

               val = d;

           next = null;

       }

   }

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

       //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

   //display() function prints linked list.

   void display(ListNode node)

   {

       //run while loop till last node.

       while (node != null)

       {

               //print node value of current node.

           System.out.print(node.val + " ");

           //move node pointer by one node.

           node = node.next;

       }

   }

   public static void main(String[] args) {

       //Create object of Linked List class.

       LinkedList list = new LinkedList();

       //first we create nodes and connect them to form a linked list.

       //Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.

       //Create a Node having node data = 1 and assign head pointer to it.

       //As head is listNode of static type. so, we call head pointer using class Name instead of object name.

       LinkedList.head = new ListNode(1);

       //Create a Node having node data = 2 and assign head.next to it.

       LinkedList.head.next = new ListNode(2);

       LinkedList.head.next.next = new ListNode(3);

       LinkedList.head.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next = new ListNode(4);

       LinkedList.head.next.next.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next.next.next = new ListNode(5);

       LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);

       //display linked list before Removing duplicates.

       System.out.println("Linked List before removing duplicates : ");

       list.display(head);

       //call removeDuplicates() function to remove duplicates from linked list.

       list.removeDuplicates();

       System.out.println("")

       //display linked list after Removing duplicates.

       System.out.println("Linked List after removing duplicates :  ");

       list.display(head);

   }

}

Output:-

Question options:

a. RAID 0

b. RAID 1

c. RAID 5

d. RAID 6

e. RAID 10

Answer:

d. RAID 6

Explanation:

RAID is Redundant Array of Inexpensive/Independent Disks. RAID combines low cost physical hard disk drives in one hard disk drive. RAID is used to achieve data redundancy(data backup but with synchronization) or improved performance or both.

To get what the director requires he would need to use RAID 6. RAID 6 is RAID level optimized to achieve data redundancy but with slow performance.

Answer:

Hence the answer is a sum.

Explanation:

Here the statement shows that the dashboard should show the sum and average as two separate columns if we name measure names because the column field then for marks we'll use the sum. Since the sum displays the sum of the values of all the fields, which refers to the marks of the person.

Therefore the answer is a sum.

If we use measure names as the column field, we should use "the sum".

A set of cells that are support equipment on about a similar vertical line, is considered as a column.

The preceding reply is thus right.

Learn more:

https://brainly.com/question/17271202

Answer:

The program in C++ is as follows:

#include <iostream>

#include <vector>

using namespace std;

int main(){

   vector<int> intVect;

   int n;

   cin>>n;

int intInp;

for (int i = 0; i < n; i++) {

 cin >> intInp;

 intVect.push_back(intInp); }

for (int i = n-1; i >=0; i--) {  cout << intVect[i] << " "; }

 return 0;

}

Explanation:

This declares the vector

   vector<int> intVect;

This declares n as integer; which represents the number of inputs

   int n;

This gets input for n

   cin>>n;

This declares intInp as integer; it is used to get input to the vector

int intInp;

This iterates from 0 to n-1

for (int i = 0; i < n; i++) {

This gets each input

 cin >> intInp;

This passes the input to the vector

 intVect.push_back(intInp); }

This iterates from n - 1 to 0; i.e. in reverse and printe the vector elements in reverse

for (int i = n-1; i >=0; i--) {  cout << intVect[i] << " "; }

Answer:

Explanation:

The following class is written in Python as it is the language that uses the init method. The class contains all of the data members as requested as well as the init and get_grade method. It also contains a setter method for both data variables and a get_name method as well. The code can also be seen in the attached image below.

class Student:

   _name = ""

   _grade = ""

   def __init__(self, name, grade):

       self._name = name

       self._grade = grade

   def set_name(self, name):

       self._name = name

   def set_grade(self, grade):

       self._grade = grade

   def get_name(self):

       return self._name

   def get_grade(self):

       return self._grade

Answer:

Here the code is given as follows,

Explanation:

class Example:

   def _init_(self, val):

       self.val = val

   def _gt_(self, other):

       return self.val > other.val

   def _sub_(self,other):

       return abs(len(self.val) - len(other.val))

def main():

   obj1 = Example('this is a string')

   obj2 = Example('this is another one')

   print(obj1 > obj2)

   print(obj1 - obj2)

main()

Output:-

Answer:

Explanation:

#include <iostream>

using std::cout;

using std::endl;  

using std::cin;  

int main()

{

int side, rowPosition, size;

cout << "Enter the square side: ";

cin >> side;

size = side;

while ( side > 0 ) {

rowPosition = size;

 while ( rowPosition > 0 ) {

if ( size == side || side == 1 || rowPosition == 1 ||  

rowPosition == size )

cout << '*';

else

cout << ' ';

--rowPosition;

}  

cout << '\n';

--side;

}

cout << endl;

return 0;

}

Answer:

Following are the code to the given question:

#include <iostream>//header file

using namespace std;

int main()//main method

{

int runTimes[] = {800, 775, 790, 805, 808};//defining array of integer

for(int x=0;x<3;x++)//defining for loop to print three element of array value

{

printf("%d\n",runTimes[x]);//print array value  

}

return 0;

}

Output:

Please find the attached file.

Explanation:

In this code, an array integer "runTimes" is declared that holds the given array values, and in the next step, a for loop is declared.

Inside the loop, an integer variable x is declared that starts from 0 and ends when its value less than 3 in which we print the first three array element values.

Answer:

Answer:In this context sound data is stored and transmitted in an analog form. Since computers represent data in digital form, (as bits and bytes) the

Answer:

Explanation:

The following is written in Java. It creates the ShoppingCart class as requested and implements the requested methods. A test class has been added to the main method and the output is highlighted in red down below.

import java.util.ArrayList;

import java.util.Scanner;

class Brainly {

   public static void main(String[] args) {

       ShoppingCart newCart = new ShoppingCart();

       newCart.add_item();

       newCart.add_item();

       newCart.add_item();

       newCart.print_summary();

   }

}

class ShoppingCart {

   Scanner in = new Scanner(System.in);

   ArrayList<String> items = new ArrayList<>();

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

   ArrayList<Double> cost = new ArrayList<>();

   public ShoppingCart() {

   }

   public void add_item() {

       System.out.println("Enter Item:");

       this.items.add(this.in.next());

       System.out.println("Enter Item Amount:");

       this.amount.add(this.in.nextInt());

       System.out.println("Enter Cost Per Item:");

       this.cost.add(this.in.nextDouble());

   }

   public void get_total_price() {

       double total = 0;

       for (double price: cost) {

           total += price;

       }

       System.out.println("Total Cost: $" + total);

   }

   public void print_summary() {

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

           System.out.println(amount.get(i) + " " + items.get(i) + " at " + cost.get(i) + " each.");

       }

       get_total_price();

   }

}

Answer and Explanation:

I and II answered

Interruption: interruption occurs when network is tampered with or communication between systems in a network is obstructed for illegitimate purposes.

Interception: interception occurs when data sent between systems is intercepted such that the message sent to another system is seen by an unauthorized user before it reaches destination. Interception violates confidentiality of a message.

Modification: modification occurs when data sent from a system to another system on the network is altered by an authorized user before it reaches it's destination. Modification attacks violate integrity, confidentiality and authenticity of a message.

Fabrication: fabrication occurs when an unauthorized user poses as a valid user and sends a fake message to a system in the network. Fabrication violates confidentiality, integrity and authenticity.

Answer:

Explanation:

The code that was provided in the question contained a couple of bugs. These bugs were fixed and the new code can be seen below as well as with the solution for the number of one-dollar bills included. The program was tested and the output can be seen in the attached image below highlighted in red.

amount_to_change = int(input())

num_fives = amount_to_change // 5

#Your solution goes here

num_ones = amount_to_change % 5

print('Change for $' + str(amount_to_change))

print(num_fives, 'five dollar bill(s) and', num_ones, 'one dollar bill (s)')

Answer:

b. 4B5B information bit rate is 80% of the total information plus overhead bit rate.

Explanation:

4B5B bit rate encoding means there is 80% of the bit rate which included all the information plus the overheads. There is 20% of lack in bit rate due to encoding code words which converts data into smaller bits.

Answer:

This is worth 50 dollars.

Explanation:

١ - one

٢ - two

٣ - three

٤ - four

٥ - five

٦ - six

٧ - seven

٨ - eight

٩ - nine

٠ - zero

What you see above is the ten digits in Arabic.

Both 5 (٥) and 0 (٠) appear here, thus representing 50.

Answer:

Which lan is it

Answer:

Đang học năm nhất trường TDT đúng k bạn :D

Explanation:

Answer and Explanation:

A spreadsheet is an interactive computer application for the analysis and storage of data. The database is a collection of data and accessed from the computer system. This is the main difference between spreadsheets and databases. The spreadsheet is accessed by the user and the database is accessed by the user. The database can store more data than a spreadsheet. The spreadsheet is used for tasks and used in enterprises to store the data. Spreadsheet and database are two methods and the main difference is the computer application that helps to arrange, manage and calculate the data. The database is a collection of data and organized to access easily. Spreadsheet applications were the first spreadsheet on the mainframe. Database stores and manipulates the data easily. The database maintains the data and stores the records of teachers and courses. The spreadsheet is the computer application and analyzing the data in table form. The spreadsheet is a standard feature for productive suite and based on an accounting worksheet.

Answer:

import math

n = int(input("Enter n: "))

for number in range(1, n+1):

   if number % 2 == 1:

       print("{:.2f}".format(math.sqrt(number)), end=" ")

Explanation:

In order to calculate the square roots of the numbers, import math. Note that math.sqrt() calculates and gives the square root of the given number as a parameter

Ask the user to enter the n

Create a for loop that iterates from 1 to n+1 (Since n is included, you need to write n+1)

Inside the loop, check whether the number is odd or not using the if structure and modulo. If the number is odd, calculate its square root using the math.sqrt() and print the result with two digits after the decimal

Answer:

Explanation:

The answer should be 1.16 or 61.1

Answer:

import java.util.Scanner;

class Main {

 public static int calcSeries(int n) {

   int sum = 0;

   for(int i=10; i>=n; i--) {

     sum += i;

   }

   return sum;

 }

 public static void main(String[] args) {

   Scanner reader = new Scanner(System.in);

   int n = 0;

   do {

     System.out.print("Enter n: ");

     n = reader.nextInt();

     if (n >= 10) {

       System.out.println("Please enter a value lower than 10.");

     }

   } while (n >= 10);

   reader.close();

   System.out.printf("sum: %d\n", calcSeries(n));

 }

}

Answer:

"TCP/IP library" is the appropriate answer.

Explanation:

Answer:

B. Parallel

Explanation:

When using the ohmmeter function of a digital multimeter, the leads are placed parallel to the component being tested. The digital multimeter is placed parallel to the component because, current has to flow into the component so as to be able to measure its resistance. Without the flow of current in the component, the resistance could not be measured.

If the component were placed in series, there would be no way to close the circuit because, we need a closed circuit so as to measure the resistance and use the ohmmeter function of the digital multimeter.

Only a parallel connection would close the circuit.

So, B is the answer.

Answer:

Explanation:

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

#define MAX 1024

int total = 0 ;

int n1, n2;

char s1, s2;

FILE fp;

int readf(FILE fp)

{

 if((fp=fopen("strings.txt", "r"))==NULL) {

   printf("ERROR: can’t open string.txt!\n");

   return 0;

 }

 s1=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 s2=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 /* read s1 s2 from the file */

 s1=fgets(s1, MAX, fp);

 s2=fgets(s2, MAX, fp);

 n1=strlen(s1); /* length of s1 */

 n2=strlen(s2)-1; /* length of s2 */

 if(s1==NULL || s2==NULL || n1<n2) /* when error exit */

   return 1;

}

int num_substring(void)

{

 int i, j, k;

 int count;

 for(i=0; i<=(n1-n2); i++) {

   count=0;

   for(j=i, k=0; k<n2; j++, k++){ /* search for the next string of size of n2 */

   if((s1+j)!=(s2+k)) {

     break;

   }

   else

     count++;

   if(count==n2)

     total++; /* find a substring in this step */

   }

 }

 return total;

}

int main(int argc, char argv[])

{

 int count;

 readf(fp);

 count=num_substring();

 printf("The number of substrings is: %d\n", count);

 return 1;

}

Answer:

5) I and III only

Explanation:

From the declarations listed both declarations I and III will not permit calling the doSomething() method. This is because in declaration I you are creating a superclass variable but initializing the subclass. In declaration III you are doing the opposite, which would work in Java but only if you cast the subclass to the superclass that you are initializing the variable with. Therefore, in these options the only viable one that will work without error will be II.

Answer:

15

Inside the type declaration, you specify the maximum length the entry can be. For branch, it would be 15.

I can't seem to type the full "vc(15)" phrase because brainly won't let me.