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lawyer [7]
3 years ago
11

Assignment 4: Evens and Odds

Computers and Technology
1 answer:
Anika [276]3 years ago
5 0

n = int(input("How many numbers do you need to check? "))

odd = 0

even = 0

i = 0

while i < n:

   num = int(input("Enter number: "))

   if num % 2 == 0:

       even += 1

       print(str(num)+" is an even number")

   else:

       odd += 1

       print(str(num)+" is an odd number")

   i += 1

print("You entered "+str(even)+" even number(s).")

print("You entered "+str(odd)+" odd number(s).")

I hope this helps!

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Drag each statement to the correct location.
VMariaS [17]

Answer:

(a)\ 222_{10} = DE_{16} --- True

(b)\ D7_{16} = 11010011_2 --- False

(c)\ 13_{16} = 19_{10} --- True

Explanation:

Required

Determine if the statements are true or not.

(a)\ 222_{10} = DE_{16}

To do this, we convert DE from base 16 to base 10 using product rule.

So, we have:

DE_{16} = D * 16^1 + E * 16^0

In hexadecimal.

D =13 \\E = 14

So, we have:

DE_{16} = 13 * 16^1 + 14 * 16^0

DE_{16} = 222_{10}

Hence:

(a) is true

(b)\ D7_{16} = 11010011_2

First, convert D7 to base 10 using product rule

D7_{16} = D * 16^1 + 7 * 16^0

D = 13

So, we have:

D7_{16} = 13 * 16^1 + 7 * 16^0

D7_{16} = 215_{10}

Next convert 215 to base 2, using division rule

215 / 2 = 107 R 1

107/2 =53 R 1

53/2 =26 R1

26/2 = 13 R 0

13/2 = 6 R 1

6/2 = 3 R 0

3/2 = 1 R 1

1/2 = 0 R1

Write the remainders from bottom to top.

D7_{16} = 11010111_2

<em>Hence (b) is false</em>

(c)\ 13_{16} = 19_{10}

Convert 13 to base 10 using product rule

13_{16} = 1 * 16^1 + 3 * 16^0

13_{16} = 19

Hence; (c) is true

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3 years ago
What is up with the bots? They are so annoying.
Yuki888 [10]

Answer:

very very very annoying

6 0
3 years ago
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Write a method reverse that takes an array as an argument and returns a new array with the elements in reversed order. Do not mo
mr Goodwill [35]

Answer:

public class ArrayUtils

{

//function to reverse the elements in given array

public static void reverse(String words[])

{

//find the length of the array

int n = words.length;

//iterate over the array up to the half

for(int i = 0;i < (int)(n/2);i++)

{

//swap the first element with last element and second element with second last element and so on.

String temp;

temp = words[i];

words[i] = words[n - i -1];

words[n - i - 1] = temp;

}

}

public static void main(String args[])

{

//create and array

String words[] = {"Apple", "Grapes", "Oranges", "Mangoes"};

//print the contents of the array

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

{

System.out.println(words[i]);

}

//call the function to reverse th array

reverse(words);

//print the contents after reversing

System.out.println("After reversing............");

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

{

System.out.println(words[i]);

}

}

Explanation:

4 0
2 years ago
The value of the mathematical constant e can be expressed as an infinite series: e=1+1/1!+1/2!+1/3!+... Write a program that app
LUCKY_DIMON [66]

Answer:

// here is code in c++ to find the approx value of "e".

#include <bits/stdc++.h>

using namespace std;

// function to find factorial of a number

double fact(int n){

double f =1.0;

// if n=0 then return 1

if(n==0)

return 1;

for(int a=1;a<=n;++a)

       f = f *a;

// return the factorial of number

return f;

}

// driver function

int main()

{

// variable

int n;

double sum=0;

cout<<"enter n:";

// read the value of n

cin>>n;

// Calculate the sum of the series

  for (int x = 0; x <= n; x++)

  {

     sum += 1.0/fact(x);

  }

  // print the approx value of "e"

    cout<<"Approx Value of e is: "<<sum<<endl;

return 0;

}

Explanation:

Read the value of "n" from user. Declare and initialize variable "sum" to store the sum of series.Create a function to Calculate the factorial of a given number. Calculate the sum of all the term of the series 1+1/1!+1/2!.....+1/n!.This will be the approx value of "e".

Output:

enter n:12

Approx Value of e is: 2.71828

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ziro4ka [17]
Cause they make up half the population of the united states of america
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3 years ago
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