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BlackZzzverrR [31]
2 years ago
11

Convert the following binary to decimal number . a). 101101 ​

Computers and Technology
2 answers:
KonstantinChe [14]2 years ago
5 0
First things first you have to do write it down which you did
step 2: then you multiply each digit of the binary number by the corresponding power of 2: 1x25 + 0x24 + 1x23 + 1x22 + 0x21 + 1x2 0
step 3: solve the powers: 1x32 + 0x16 + 1x8 + 1x4 + 0x2 + 1x1 = 32 + 0 + 8 + 4 + 0 + 1
step 4: add the numbers written above

32 + 0 + 8 + 4 + 0 + 1 = 45.
So, 45 is the decimal equivalent of the binary number 101101.
adoni [48]2 years ago
3 0

(101101)_2\\\\=1\times 2^5 +  0 \times 2^4 + 1 \times 2^3 +1 \times 2^2 + 0 \times 2^1+1 \times 2^0\\\\=32+0+8+4+0+1\\\\=(45)_{10}

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Answer:

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Create an application containing an array that stores eight integers. The application should call five methods that in turn (1)
Butoxors [25]

Answer:

package b4;

public class ArrayMethodDemo {

public static void main(String[] args) {

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 int[] integernumbers = { 3, 4, 6, 9, 5, 6, 7, 2 };

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 // elements.

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}

// Method to display the elements in the array.

// The method receives only a single argument which is the array.

// Loop through the array and at every cycle, print out each element

public static void display(int[] arr) {

 System.out.print("All integers in the array : ");

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

  System.out.print(arr[i] + " ");

 }

 System.out.println();

}

// Method to display the elements in the array in reverse order

// The method receives only a single parameter which is the array

// Loop through the array starting at the last index

// At every cycle, print out the elements in the array

public static void displayReverse(int[] arr) {

 System.out.print("All integers in the array in reverse : ");

 for (int i = arr.length - 1; i >= 0; i--) {

  System.out.print(arr[i] + " ");

 }

 System.out.println();

}

// Method to print out the sum of the elements in the array.

// The method receives only a single parameter which is the array.

// Declare a variable called sum to hold the sum of the elements

// Initialize sum to zero

// Loop through the array and cumulatively add each element to sum

// Print out the sum at the end of the loop

public static void sum(int[] arr) {

 int sum = 0;

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

  sum += arr[i];

 }

 System.out.println("The sum of the integers in the array is " + sum);

}

// Method to print all values in the array that are less than a limiting

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// The method has two parameters - the array and the limiting argument.

// Loop through the array and at every cycle,

// check if the current array element is less than the limiting argument.

// If it is, print it out. Else continue

public static void lessThan(int[] arr, int limitingargument) {

 System.out.print("All values less than the limiting argument (" + limitingargument + ") are: ");

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

  if (arr[i] < limitingargument) {

   System.out.print(arr[i] + " ");

  }

 }

 System.out.println();

}

// Method to print all values in the array that are higher than the average of

// the array.

// The method has one parameter - the array.

// First, calculate the average of the array elements.

// Loop through the array and at every cycle,

// check if the current array element is greater than the average.

// If it is, print it out. Else continue

public static void higherThan(int[] arr) {

 int sum = 0;

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

  sum += arr[i];

 }

 double average = sum / arr.length;

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

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  }

 }

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}

}

Explanation:

The program has been written in Java.

Please go through the comments in the code for explanation.

The source code has also been attached to this response.

Hope this helps!

Download java
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3 years ago
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mamaluj [8]
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3 years ago
2.8 Code Practice: Question 1
yuradex [85]

Answer:

Explanation:

First take the input of the numbers:

x = int(input("Enter the number of elements to be in the list:"))

b=[ ]

for i in range(0,x):

a=int(input("Element: "))

b.append(a)

c=[ ]

d=[ ]

for i in b:

if(i%2==0):

c.append(i)

else:

d.append(i)

c.sort()

d.sort()

count1=0

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for k in c:

count1=count1+1

for j in d:

count2=count2+1

print("Largest even number:",c[count1-1])

print("Largest odd number",d[count2-1])

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3 years ago
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