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ruslelena [56]
3 years ago
9

Mary is a big fan of tropical fish. She has a few tanks of fish at home. To maintain a healthy environment for the fish, she nee

ds to add conditioner to the water once a week. The amount of conditioner added is determined by the volume of the tank. According to the direction on the bottle, she has to add 1 ml of conditioner per 100 cubic inches of water. She wants a program to calculate exactly how much conditioner to add to a tank of water. All tanks are in rectangular shape. The program will ask for the length, the width and the height of the tank. It will calculate and display the amount of conditioner to be added. [Note: volume = length X width X height] Use the following two test cases for desk-checking [all in inches]:
length width height
20 22.5 10
16 10 15
Write a C++ programs that implements and then add code to write the result to an output file. Make sure you generate a description message with the data in the output file (i.e. The amount of conditioner to be added is 3 ml).
outFile << "The amount of conditioner to be added is "<< amount<<" ml"< Name output data file "conditioner.txt" .
Computers and Technology
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

Answered below

Explanation:

# Program is written in Python programming language

conditioner_in_ml = 0

width = float(input("Enter width in inches: "))

height = float(input("Enter height in inches: "))

length = float (input("Enter length in inches: "))

#Calculate the volume

volume = width * length * height

#Calculate the amount of conditioner per 100 #cubic inches of volume

conditioner_in_ml = volume/ 100

print("The amount of conditioner required for $volume cubic inches is $conditioner_in_ml ml")

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Python - Write a program to print the multiplication table as shown in the image by using for loops.
Galina-37 [17]

Answer:

Explanation:

The following python code creates the multiplication table for 10 rows and 10 columns. This code uses nested for loops to traverse the table and print out the product of each multiplication. The image attached shows the output of the code.

for x in range(1, 11):

       for y in range(1, 11):

           z = x * y

           print(z, end="\t")

       print()

8 0
3 years ago
Question 6 options: This is a category or partitioned group of the small units of a programming language include its numeric lit
Wittaler [7]

Lexeme is a category or partitioned group of the small units of a programming language include its numeric literals, operators, and special words.

<h3>What is a lexeme in programming?</h3>

Lexemes are known to be those character strings that are put together from the character group of a program, and the token stands for what aspect of the program's grammar they are made of.

Hence, Lexeme is a category or partitioned group of the small units of a programming language include its numeric literals, operators, and special words.

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6 0
2 years ago
Create a Java program with threads that looks through a vary large array (100,000,000 elements) to find the smallest number in t
olchik [2.2K]

Answer:

See explaination

Explanation:

import java.util.Random;

public class Sample{

static class MinMax implements Runnable{

int []arr;

int start,end,min,max;

MinMax(int[]arr, int start,int end){

this.start=start;

this.end=end;

min=Integer.MAX_VALUE;

max=Integer.MIN_VALUE;

this.arr=arr;

}

atOverride

public void run() {

for(int i=start;i<=end;i++){ //search min and max form strant to end index

min=Math.min(min,arr[i]);

max=Math.max(max, arr[i]);

}

}

}

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

long beginTime = System.nanoTime();

Random gen = new Random();

int n=100000000;

int[] data = new int[n]; //generate and fill random numbers

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

data[i] = gen.nextInt()%1000000;

}

long endTime = System.nanoTime();

System.out.println("Done filling the array. That took " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 1 thread"); //1 thread

MinMax m1=new MinMax(data,0,n-1); //class object

Thread t1=new Thread(m1); //new thread

beginTime=System.nanoTime(); //start timer

t1.start(); //start thread

t1.join(0); //wait until thread finishes

endTime=System.nanoTime(); //end timer

System.out.println("Min,Max: "+m1.min+","+m1.max); //print minimum and maximum

System.out.println("Time using 1 thread " + (endTime - beginTime)/1000000000f + " seconds."); //print time taken

//-----------------------------------------

System.out.println("Using 2 thread");

m1=new MinMax(data,0,n/2);

MinMax m2=new MinMax(data,n/2+1,n-1);

t1=new Thread(m1);

Thread t2=new Thread(m2);

beginTime=System.nanoTime();

t1.start();

t2.start();

t1.join(0);

t2.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(m1.min,m2.min)+","+Math.max(m1.max,m2.max));

System.out.println("Time using 2 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 3 thread");

m1=new MinMax(data,0,n/3);

m2=new MinMax(data,n/3+1,2*n/3);

MinMax m3=new MinMax(data,2*n/3+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

Thread t3=new Thread(m3);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t1.join(0);

t2.join(0);

t3.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),m3.min)+","+Math.max(Math.max(m1.max,m2.max),m3.max));

System.out.println("Time using 3 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 4 thread");

m1=new MinMax(data,0,n/4);

m2=new MinMax(data,n/4+1,2*n/4);

m3=new MinMax(data,2*n/4+1,3*n/4);

MinMax m4=new MinMax(data,3*n/4+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

t3=new Thread(m3);

Thread t4=new Thread(m4);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t4.start();

t1.join(0);

t2.join(0);

t3.join(0);

t4.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),Math.min(m3.min,m4.min))+","+Math.max(Math.max(m1.max,m2.max),Math.max(m3.max,m4.max)));

System.out.println("Time using 4 thread " + (endTime - beginTime)/1000000000f + " seconds.");

}

}

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3 years ago
Assume that you have an ArrayList variable named a containing 4 elements, and an object named element that is the correct type t
jonny [76]

Answer:

Option (4) is the correct answer.

Explanation:

In Java programming language ,array collection starts from 0 index location and ends in a size-1 index location. So to access the last elements the user needs to use a[Size-1] statement. so to modify the value of the last location of the array the user needs to use "a[size-1]= element;".

But when the user wants to add some new value to the end of the array list collection then he needs to use the statement--

a.add(element); //where add is a function, element is a value and a is a array list object.

Another option is invalid because--

  • Option 1 is not the correct because "a[3]=element;" modify the value of the 3rd element of the array.
  • Option 2 gives a compile-time error because add functions bracts are not closed.
  • Option 3 gives the error because a[4] gives the location of the 5th element of the array but the above question says that a is defined with 4 elements.
7 0
3 years ago
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