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3 years ago

What is the benefit of the load balancing logic to end-user?

Computers and Technology

1 answer:

Alborosie3 years ago

6 0

Answer:

<em>It can minimize response time, and minimize the costs for the end user.</em>

Explanation

Load balancing is a form of workload distribution across several computers or resources. It allows each segment of the system to process a smaller task, minimizing respond time, helping to avoid overload, and contributing to optimize resource use.

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A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter

trapecia [35]

Answer:

5 Letters

Explanation:

So we need 12 distinct codes made of a single letter or a pair of letters.

What would be the least number of letters?

Lets try with 3 letters A, B and C

The possible combinations are: A, B, C, AB, AC, BC

These are 6 codes and we need 12 so lets try more A, B, C and D

The possible combinations are: A, B, C, D, AB, AC, AD, BC, BD, CD

These are 10 codes and we need 12 so lets try more A, B, C, D and E

The possible combinations are:

A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

Finally we got 15 distinct codes which are more than 12 so the least number of letters needed are 5.

Using formula:

Four letters = 4C1 + 4C2 = 4 + 6 = 10

Five letters = 5C1 + 5C2 = 5 + 10 = 15

5 0

3 years ago

Write a program whose input is two integers. Output the first integer and subsequent increments of 5 as long as the value is les

adell [148]

Answer:

Following are the code to the given question:

For code 1:

start = int(input())#defining a start variable that takes input from the user end

end = int(input())#defining a end variable that takes input from the user end

if start > end:#use if that checks start value greater than end value

print("Second integer can't be less than the first.")#print message

else:#defining else block

while start <= end:#defining a while loop that checks start value less than equal to end value

print(start, end=' ')#print input value

start += 5#incrementing the start value by 5

print()#use print for space

For code 2:

while True:#defining a while loop that runs when its true

data = input()#defining a data variable that inputs values

if data == 'Done' or data == 'done' or data == 'd':#defining if block that checks data value

break#use break keyword

rev = ''#defining a string variable rev

for ch in data:#defining a for loop that adds value in string variable

rev = ch + rev#adding value in rev variable

print(rev)#print rev value

Explanation:

In the first code two-variable "first and end" is declared that takes input from the user end. After inputting the value if a block is used that checks start value greater than end value and use the print method that prints message.

In the else block a while loop is declared that checks start value less than equal to end value and inside the loop it prints input value and increments the start value by 5.

In the second code, a while loop runs when it's true and defines a data variable that inputs values. In the if block is used that checks data value and use break keyword.

In the next step, "rev" as a string variable is declared that uses the for loop that adds value in its variable and prints its values.

3 0

3 years ago

What data unit is encapsulated inside a packet?

viva [34]

Answer: datagram

Explanation:

7 0

3 years ago

Read 2 more answers

Input 10 integers and display the following:

LekaFEV [45]

Answer:

// code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variables

int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;

int largest=INT_MIN;

int smallest=INT_MAX;

int n;

cout<<"Enter 10 Integers:";

// read 10 Integers

for(int a=0;a<10;a++)

{

cin>>n;

// find largest

if(n>largest)

largest=n;

// find smallest

if(n<smallest)

smallest=n;

// if input is even

if(n%2==0)

{

// sum of even

sum_even+=n;

// even count

eve_count++;

}

else

{

// sum of odd

sum_odd+=n;

// odd count

odd_count++;

}

// print sum of even

cout<<"Sum of all even numbers is: "<<sum_even<<endl;

// print sum of odd

cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;

// print largest

cout<<"largest Integer is: "<<largest<<endl;

// print smallest

cout<<"smallest Integer is: "<<smallest<<endl;

// print even count

cout<<"count of even number is: "<<eve_count<<endl;

// print odd cout

cout<<"count of odd number is: "<<odd_count<<endl;

return 0;

}

Explanation:

Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.

Output:

Enter 10 Integers:1 3 4 2 10 11 12 44 5 20

Sum of all even numbers is: 92

Sum of all odd numbers is: 20

largest Integer is: 44

smallest Integer is: 1

count of even number is: 6

count of odd number is: 4

3 0

3 years ago

How has the development of personal computer hardware and software reversed some of the trends brought on by the industrial revo

Nadya [2.5K]

<span>There should be no need to have white-collar employers just next to factories thanks to the Information Revolution, and it enabled executives to stay in home, write documents, and send them easily.</span>

4 0

4 years ago