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marin [14]
3 years ago
7

Write an algorithm that accepts two numbers,

Computers and Technology
1 answer:
Paladinen [302]3 years ago
3 0
Let’s write the algorithm in the form of a pseudocode!

Pseudocode:Quotient_of_Two_Number
Declare: num1, num2, quotient
START
Display (Enter a number)
Read num1
Display (Enter a number)
Read num2
quotient = num1/num2
Print (“Quotient”)
STOP
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What is the difference between a loop and a function?
Eduardwww [97]
Big difference
Loops allow you to execute code multiple times while a condition is true
Functions allow you to “call” a snippet of code whenever you want, and you can pass it arguments that could affect the data it returns
7 0
2 years ago
Read 2 more answers
Given an array as follows
slava [35]

Answer:

1) Method calcTotal:

  1. public static long calcTotal(long [][] arr2D){
  2.        long total = 0;
  3.        
  4.        for(int i = 0; i < arr2D.length; i++ )
  5.        {
  6.            for(int j = 0; j < arr2D[i].length; j++)
  7.            {
  8.                total = total + arr2D[i][j];
  9.            }
  10.        }
  11.        
  12.        return total;
  13.    }

Explanation:

Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

Line 2: Declare a variable, total, and initialize it with zero.

Line 4: An outer for-loop to iterate through every rows of the two-dimensional array

Line 6: An inner  for-loop to iterate though every columns within a particular row.

Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

Line 12: Return the final total of all the element values as an output

Answer:

2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
  16.        
  17.        return average;
  18.    }

Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

3) calcRowAverage:

  1. public static double calcRowAverage(long [][] arr2D, int row){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            if(i == row)
  8.            {
  9.                for(int j = 0; j < arr2D[i].length; j++)
  10.                {
  11.                    total = total + arr2D[i][j];
  12.                    count++;
  13.                }
  14.            }
  15.            
  16.        }
  17.        
  18.        double average = total / count;
  19.        
  20.        return average;
  21.    }

Explanation:

By using method <em>calcAverage </em>as a foundation, add one more parameter, row, in the method <em>calcRowAverage</em>. The row number is used as an conditional checking criteria to ensure only that particular row of elements will be summed up and divided by the counter to get an average of that row.

Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

5 0
3 years ago
Consider sending a 10000-byte datagram into a link that has an MTU of 4468 bytes. Suppose the original datagram is stamped with
AfilCa [17]

Answer:

Number of fragments is 3

Explanation:

The maximum size of data field in each fragment = 4468 - 20(IP Header)

= 4448 bytes

Hence, the number of required fragment = (10000 - 20)/4448

= 3

Fragment 1

Id = 218

offset = 0

total length = 4468 bytes

flag = 1

Fragment 2

Id = 218

offset = 556

total length = 4468 bytes

flag = 1

Fragment 3

Id = 218

offset = 1112

total length = 1144 bytes

flag = 0

5 0
3 years ago
Please help it's my last question
Yuki888 [10]

Explanation:

here is your answer.. of. different between client / server architecture and peer to peer architecture of the network.

6 0
3 years ago
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Mary has cleaned her data and is ready to determine the most efficient bus route. She starts by splitting the city into four reg
igomit [66]

ANSWER:

B.

Transforming the data might help Mary notice a different pattern that makes a bigger impact on bus routes than regions of the City.

EXPLANATION:

Mary can check if there are other things or factors that might influence where the bus routes need to be prioritized. For example if the students' age is considered for the bus routes instead of regions. She can achieve this by

By Re-sorting or Transforming the data.

That is to say she will be able to find out if it is only region or if there are different patterns that makes bigger impacts on bus routes, by transforming the data.

8 0
3 years ago
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