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slava [35]
3 years ago
14

Write a program that uses a 2-D array to store the highest and lowest temperatures for each month of the year. The program shoul

d output the average high, average low, and highest and lowest temperatures of the year. Your program must consist of the following methods with their appropriate parameters:
Computers and Technology
1 answer:
dedylja [7]3 years ago
7 0

Answer:

*/

import java.util.*;

public class Main{

public static void main(String[] args){

 

int infoArray[][]=getData();

int avgArray[][]=average(infoArray);

int tempArray[][]=temps(infoArray);

System.out.println("Data set - Highs: 87,99,100,102,101,97,97,93,99,94,98,92.");  

System.out.println("Lows: 11,12,19,24,23,25,32,33,29,21,10,1.");

System.out.println("Average High: "+avgArray[0][0]);

System.out.println("Average Low: "+avgArray[1][0]);

System.out.println("Highest Temperature: "+tempArray[0][0]);

System.out.println("Lowest Temperature: "+tempArray[1][0]);

}

public static int[][] getData(){

 int tempDatahl[][] = {{87,99,100,102,101,97,97,93,99,94,98,92},{11,12,19,24,23,25,32,33,29,21,10,1}};

 return tempDatahl;

}

public static int[][] average(int x[][]){

int avgH = 0;

int avgL = 0;

 

//loop to find avg high

 for(int i=0;i<x[0].length;i++){

     avgH+=x[0][i]  ;

}

 

//loop to find avg low

 for(int h=0;h<x[1].length;h++){

     avgL+=x[1][h]  ;

}  

 

 avgH=avgH/x[0].length;

 avgL=avgL/x[1].length;

 

 int avgData[][] = {{avgH},{avgL}};

 return avgData;

}

public static int[][] temps(int x[][]){

 //high temp

 int ht = 0;

 int lt = 100;

 

 for(int t=0;t<x[0].length;t++){

   if(x[0][t]>ht){

     ht = x[0][t];

   }

 }

 //low temp

  for(int y=0;y<x[1].length;y++){

   if(x[1][y]<lt){

     lt = x[1][y];

   }

 }  

      int arrTemps[][]={{ht},{lt}};

     return arrTemps;

}

 

}

Explanation:

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stepan [7]

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  • Re-balance Work and Home.
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hope it helps! :)

3 0
3 years ago
Complete the sentence.
mamaluj [8]

Answer:

We need context, it could be several answers

Explanation:

4 0
3 years ago
Given that k refers to a non-negative int value and that t has been defined and refers to a tuple with at least k+1 elements, wr
Rus_ich [418]

Answer:

The expression is "t[k]".

Explanation:

In the question, it is defined that variable k refers to positive integer and variable t refers to a tuple that holds at least k+1 element. The expression for defining the kth element of t is t[k].

  • Where variable t is tuple that collects a sequence of immutable (in-built) Python objects. To define tuple we use the parentheses.
  • The main purpose to use tuple in this question is that it cannot be changed.  

5 0
3 years ago
You are a visitor at a political convention with delegates; each delegate is a member of exactly one political party. It is impo
Fed [463]

Answer:

The algorithm is as follows:

Step 1: Start

Step 2: Parties = [All delegates in the party]

Step 3: Lent = Count(Parties)

Step 4: Individual = 0

Step 5: Index = 1

Step 6: For I in Lent:

Step 6.1: If Parties[Individual] == Parties[I]:

Step 6.1.1: Index = Index + 1

Step 6.2: Else:

Step 6.2.1 If Index == 0:

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

Step 7: Else

Step 7.1: Index = Index - 1

Step 8: Print(Party[Individual])

Step 9: Stop

Explanation:

The algorithm begins here

Step 1: Start

This gets the political parties as a list

Step 2: Parties = [All delegates in the party]

This counts the number of delegates i.e. the length of the list

Step 3: Lent = Count(Parties)

This initializes the first individual you come in contact with, to delegate 0 [list index begins from 0]

Step 4: Individual = 0

The next person on the list is set to index 1

Step 5: Index = 1

This begins an iteration

Step 6: For I in Lent:

If Parties[Individual] greets, shakes or smile to Party[i]

Step 6.1: If Parties[Individual] == Parties[I]:

Then they belong to the same party. Increment count by 1

Step 6.1.1: Index = Index + 1

If otherwise

Step 6.2: Else:

This checks if the first person is still in check

Step 6.2.1 If Index == 0:

If yes, the iteration is shifted up

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

If the first person is not being checked

Step 7: Else

The index is reduced by 1

Step 7.1: Index = Index - 1

This prints the highest occurrence party

Step 8: Print(Party[Individual])

This ends the algorithm

Step 9: Stop

The algorithm, implemented in Python is added as an attachment

<em>Because there is an iteration which performs repetitive operation, the algorithm running time is: O(n) </em>

Download txt
5 0
3 years ago
JAVA Code:
Burka [1]
Dear ,you should ask it on stackoverflow and geekforgeelks , not here
6 0
3 years ago
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