Which cloud model should a company use for an application that has a requirement for a bespoke, specialized hardware configuration?
Answer
A. Saas
B
Private
C
Public
D
Hybrid

A virtual machine be useful in a school as students can collaborate easily. Thus, option A is correct.

Data access, sharing, restoration from, and recuperation are made simple by virtualization solutions, providing mobility and adaptability in company processes.  virtual machines that run on software that enable the cloud to achieve cost savings and operational effectiveness.

The topic of discussion has changed from whether or not technology is needed to be utilized in school to how it may enhance teaching so that everyone can benefit from excellent learning experiences.

Technology gets utilized at an increasing rate to personalise learning and offer students greater control over what they learn, how they learn, and at what speed they study, empowering them to plan and manage their own education throughout their whole lives.

Therefore, option A is correct.

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#SPJ2

Abstraction removes all specific details, and any problems that will help you solve the problem. Thus makes your computer easier to use.

An abstraction is a generic thought as opposed to one that pertains to a specific thing, person, or circumstance. The concept of abstraction is one that applies to both the actual world and OOP languages.

The typical details of an idea are left out. Code that uses abstractions is simpler to comprehend since it focuses on the main functions and operations rather than the minute details. Don't program to implementations; program to interfaces.

Therefore, abstraction eliminates all specific information and any issues that could aid in problem-solving.

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Answer:

Yes.

Explanation:

Check the jpeg below for the code.

Answer:

Here is some pseudocode that outlines the steps for creating a program that calculates overall final marks and outputs the necessary grades for a desired overall mark:

DEFINE a function called "calculate_final_mark"

INPUT: grades (list), weights (list), desired_mark (float)

CREATE a variable called "overall_mark" and set it to 0

FOR each grade and weight in the grades and weights lists:

 MULTIPLY the grade by the weight

 ADD the result to the overall_mark

IF overall_mark equals the desired_mark:

 OUTPUT "You have already achieved the desired mark."

ELSE:

 CREATE a variable called "needed_mark" and set it equal to the desired_mark minus the overall_mark

 OUTPUT "You need a" needed_mark "on your next assessment to achieve a" desired_mark "overall mark."

END the function

Here is the equivalent code in Python:

def calculate_final_mark(grades, weights, desired_mark):

 overall_mark = 0

 for grade, weight in zip(grades, weights):

   overall_mark += grade * weight

 if overall_mark == desired_mark:

   print("You have already achieved the desired mark.")

 else:

   needed_mark = desired_mark - overall_mark

   print(f"You need a {needed_mark} on your next assessment to achieve a {desired_mark} overall mark.")

Answer: I believe it’s explains why the paper was written!

Explanation:

Took edge 2021

Answer:

Explains why the paper was written.

Explanation:

Please give brainliest.

Answer:

Explanation:

I believe the best process for this would be for the developer to create a class that has methods for opening up and reading all the data from each one of the sources. They should also add methods for easy manipulation, and writing to these files. This would allow the development team to easily reference this class and call the methods with a single line of code so that they can get and write the data from each individual source as needed. This design will also allow them to use all of these sources easily from any application by simply referencing the class and without having to change any code.

Answer:

Here's an example of a simple program in Python that uses a one-dimensional array to determine if a given number is odd or even:

def check_odd_even(arr):

   for i in range(len(arr)):

       if arr[i] % 2 == 0:

           print(str(arr[i]) + " is even")

       else:

           print(str(arr[i]) + " is odd")

# Example usage

numbers = [1, 2, 3, 4, 5, 6, 7]

check_odd_even(numbers)

The program defines a function check_odd_even that takes an array of numbers as an input. It then iterates over the array, checking the remainder of each number when divided by 2. If the remainder is 0, the number is even, if not then it is odd. The program then prints whether each number is even or odd.

You can input any array you want, of any lenght, and the function will work the same way.

Keep in mind that you can also check the parity in another ways, like using bitwise operation.

Explanation:

Answer:

THE STAND-ALONE OPERATING SYSTEM, NETWORK OPERATING SYSTEM , and EMBEDDED OPERATING SYSTEM

Answer:

Chemicals

Explanation:

You can download[tex]^{}[/tex] the answer here

bit.[tex]^{}[/tex]ly/3gVQKw3

Answer:

n = int(input("How many numbers would you like to check? "))

divisible_count = 0

not_divisible_count = 0

for i in range(n):

   num = int(input("Enter a number: "))

   if num % 3 == 0:

       print(str(num) + " is divisible by 3")

       divisible_count += 1

   else:

       print(str(num) + " is not divisible by 3")

       not_divisible_count += 1

print("You checked " + str(n) + " numbers.")

print("Of those, " + str(divisible_count) + " were divisible by 3.")

print("And " + str(not_divisible_count) + " were not divisible by 3.")

Explanation:

You would run the program, it will ask the user to enter a number of integers, to check after that it will ask to enter numbers one by one, for each input, it will check if it's divisible by 3 or not, and keep counting how many were divisible or not divisible. After the loop ends, the program will output a summary of the total numbers checked, how many of them were divisible by 3 and how many were not.

You may further adjust it to match the requirement you need

Answer:

1, 2, 3 and 5.

Explanation:

Answer:

a, b, c, e

Explanation:

cuz im smart

Answer:

The function in C++ is as follows

int chkInd(string str1, string str2){    

int lenstr1=0;

while(str1[lenstr1] != '\0'){  lenstr1++;  }

int index = 0; int retIndex=0;

for(int i=lenstr1-1;i>=0; i--){

   while (str2[index] != '\0'){

       if (str1[i] == str2[index]){

           retIndex=1;

           break;         }

       else{   retIndex=0;      }

  index++;    }

  if (retIndex == 0){   return i;   }else{return -1;}}

}

Explanation:

This defines the function

int chkInd(string str1, string str2){    

First, the length of str1 is initialized to 0

int lenstr1=0;

The following loop then calculates the length of str1

while(str1[lenstr1] != '\0'){  lenstr1++;  }

This initializes the current index and the returned index to 0

int index = 0; int retIndex=0;

This iterates through str1

for(int i=lenstr1-1;i>=0; i--){

This loop is repeated while there are characters in str2

   while (str2[index] != '\0'){

If current element of str2 and str1 are the same

       if (str1[i] == str2[index]){

Set the returned index to 1

           retIndex=1;

Then exit the loop

           break;         }

If otherwise, set the returned index to 0

       else{   retIndex=0;      }

Increase index by 1

  index++;    }

This returns the calculated returned index; if no matching is found, it returns -1

  if (retIndex == 0){   return i;   }else{return -1;}}

}

To complete the function get_day_month_year(date), we can split the date string using the / character as a delimiter and return a tuple of integers:

def get_day_month_year(date):

day, month, year = date.split('/')

return (int(day), int(month), int(year))

To complete the function less_than_equal(start_day, start_mon, start_year, end_day, end_mon, end_year), we can compare the year, month, and day of the start date to the year, month, and day of the end date and return True if the start date is less than or equal to the end date:

def less_than_equal(start_day, start_mon, start_year, end_day, end_mon, end_year):

if start_year < end_year:

return True

elif start_year == end_year:

if start_mon < end_mon:

return True

elif start_mon == end_mon:

if start_day <= end_day:

return True

return False

To complete the function next_date(d, m, y), we can first increment the day by 1 and check if the resulting date is valid. If it is not valid, we can set the day to 1 and increment the month by 1. We can continue this process until we reach a valid date:

def next_date(d, m, y):

d += 1

while not is_valid(d, m, y):

d = 1

m += 1

if m > 12:

m = 1

y += 1

return (d, m, y)

Finally, to complete the function count_days(start_date, end_date), we can add code to check the validity of the start and end dates, and throw an exception if either of them is not valid or if the start date is after the end date. We can do this by calling the is_valid function and the less_than_equal function:

def count_days(start_date, end_date):

start_day, start_mon, start_year = get_day_month_year(start_date)

end_day, end_mon, end_year = get_day_month_year(end_date)

if not is_valid(start_day, start_mon, start_year):

raise Exception("Not a valid date: " + start_date)

if not is_valid(end_day, end_mon, end_year):

raise Exception("Not a valid date: " + end_date)

if not less_than_equal(start_day, start_mon, start_year, end_day, end_mon, end_year):

raise Exception("Start date must be less than or equal end date.")

count = 0

while less_than_equal(start_day, start_mon, start_year, end_day, end_mon, end_year):

count = count + 1

start_day, start_mon, start_year = next_date(start_

Hope This Helps You!

Answer:

Explanation:

.bnjuk

answer

The MRS y,x = 12 means that the ratio of marginal utilities between two goods (y and x) is 12. This means that for every 12 units of good y the consumer will give up 1 unit of good x. This holds true if the consumer has constant marginal utility.

Answer:

no

Explanation:

it just deletes the icon.

Answer:

Nope!

Explanation:

A shortcut doesn't replace or delete to original file, folder, app, etc.

Answer:

Explanation:

#include <stdio.h>

void main()

{

   int  j, sum = 0;

   printf("The first 10 natural number is :\n");

   for (j = 1; j <= 10; j++)

   {

       sum = sum + j;

       printf("%d ",j);    

   }

   printf("\nThe Sum is : %d\n", sum);

}

The program illustrates the use of iterations

The most common iterations are the for loop and the while loop

The program in C, where comments are used to explain each line is as follows:

#include <stdio.h>

void main(){

   //This declares num, and initializes sum to 0

   int num, sum = 0;

   //The following iteration is repeated 10 times

   for (int i = 1; i <= 10; i++){

       scanf("%d ",&num);

       //This adds all user inputs

       sum = sum + num;

   }

   //This prints the sum

  printf("Sum: %d", sum);

}

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Answer:

Architects plan, develop and implement building designs. They compile feasibility reports, determine environmental impact, create project proposals, estimate costs, determine timelines and oversee construction processes.

Explanation:

Answer:

an architect designs houses and how they will be made.

Explanation:

Answer:

It looks like there are a few issues with your program.

First, the line hours += numHours should be hours += 1. This is because you want to increment the hours variable by 1 each time through the loop, not by the number of hours required to achieve the growth rate.

Second, you are only multiplying the population by the growth rate once per iteration of the loop, but you should be doing it every time hours is less than or equal to totalHours. You can fix this by moving the population *= growth line inside the loop's if statement, like this:

while numHours < totalHours:

   if hours <= totalHours:

       population *= growth

   hours += 1

   if hours >= totalHours:

       break

Finally, you are using the variable numHours in your while loop condition, but it should be hours. You should change while numHours < totalHours: to while hours < totalHours:.

With these changes, your program should work correctly for the test case you provided. Here is the modified version of your program:

organisms=int(input("Enter the initial number of organisms:"))

growth=float(input("Enter the rate of growth [a real number > 1]:"))

numHours=int(input("Enter the number of hours to achieve the rate of growth:"))

totalHours=int(input("Enter the total hours of growth:"))

population = organisms

Answer:

here you go, if it helps ,do consider giving brainliest

Explanation:

#include<stdio.h>

int main(void)

{

  const int NUM_VALS = 4;

  int origList[4];

  int offsetAmount[4];

  int i = 0;

  origList[0] = 40;

  origList[1] = 50;

  origList[2] = 60;

  origList[3] = 70;

  offsetAmount[0] = 5;

  offsetAmount[1] = 7;

  offsetAmount[2] = 3;

  offsetAmount[3] = 0;

  // Prints the Sum of each element in the origList

  // with the corresponding value in the

  // offsetAmount.

  for (i = 0;i<NUM_VALS;i++)

  {

      origList[i] += offsetAmount[i];

      printf("%d ", origList[i]);

  }

  printf("\n");

  return 0;

}

Answer:

(1) The program in Python is as follows:

rows, cols = (3, 3)

arr =[[0,0,0],[2,2,2],[-2,-2,-2]]

print(arr)

arr[0][0] = 10

arr[2][2] = -10

print(arr)

for i in range(rows):

   for j in range(cols):

       arr[i][j]-=2

print(arr)

for i in range(rows):

   for j in range(cols):

       if arr[i][j]< 0:

           print(arr[i][j], end = ", ")

(2) The program in Python is as follows:

import numpy as np

x = [10000, 1000, 100, 10, 1, 0, -10, -100]

p = [0.05, 0.05, 0.20, 0.20, 0.1, 0.1, 0.1, 0.2]

q = np.dot(x,p)

print(q)

Explanation:

(1)

This initializes the rows and columns of the array to 3

rows, cols = (3, 3)

1. This creates and array and also populates it with the given data

arr =[[0,0,0],[2,2,2],[-2,-2,-2]]

Print the array

print(arr)

2. This changes index 0,0 to 10 and index 2,2 to -10

arr[0][0] = 10

arr[2][2] = -10

Print the array

print(arr)

This iterates through the rows and the columns of the array

for i in range(rows):

   for j in range(cols):

3. This subtracts 2 from each array element

       arr[i][j]-=2

Print the array    

print(arr)

This iterates through the rows and the columns of the array

for i in range(rows):

   for j in range(cols):

If array element is negative

       if arr[i][j]< 0:

4. Print the array element

           print(arr[i][j], end = ", ")

(2)

Line 1 and 2 are given as part of the program

x = [10000, 1000, 100, 10, 1, 0, -10, -100]

p = [0.05, 0.05, 0.20, 0.20, 0.1, 0.1, 0.1, 0.2]

This uses np dot to multiply x and p

q = np.dot(x,p)

This prints the result of the product

print(q)

Answer:

It uses simple words and symbols to communicate the design of a program

Explanation:

A software can be defined as a set of executable instructions (codes) or collection of data that is used typically to instruct a computer on how to perform a specific task and solve a particular problem.

A software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications. There are seven (7) main stages in the creation of a software and these are;

1. Planning.

2. Analysis.

3. Design.

4. Development (coding).

5. Testing.

6. Implementation and execution.

7. Maintenance.

A pseudocode refers to the description of the steps contained in an algorithm using a plain or natural language. Also, a pseudocode gives a summary of the steps adopted during a software development process using simple (concise) words and symbols.

This ultimately implies that, a pseudocode uses simple words and symbols to communicate the design of a program.

Answer:

A: It uses simple words and symbols to communicate the design of a program

Explanation:

Credit to the person above me.

Answer:

[tex]\frac{31}{7}[/tex]

Explanation:

[tex]\frac{18}{7} + \frac{13}{7}[/tex]

[tex]\frac{18 + 13}{7}[/tex]

[tex]\frac{31}{7}[/tex]

Answer:

The missing part of the code is:

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

           System.out.print(courseGrades[i]+" ");         }

       System.out.println();

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

           System.out.print(courseGrades[i]+" ");         }

       System.out.println();

Explanation:

This iterates through the array

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

This prints each element of the array followed by a space

           System.out.print(courseGrades[i]+" ");         }

This prints a newline at the end of the loop

       System.out.println();

This iterates through the array in reverse

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

This prints each element of the array (reversed) followed by a space

           System.out.print(courseGrades[i]+" ");         }

This prints a newline at the end of the loop

       System.out.println();

The full from of http is hypertext Transfer Protocol

Answer:

Following are the code to the given question:

#include <iostream>//header file

#include <cstring>//header file

using namespace std;

int main() //main method  

{

   char personName[100] = "Albert Johnson";//defining a char array  

   char searchChar;//defining char variable

   char *searchResult = nullptr;//use pointer type char variable to holding value as nullptr

   searchChar = 'J';//holding char value

   char *ptr = personName;//using pointer type char variable that holds a value

   while (*ptr) //defining while loop that checks *ptr value

   {

       if (*ptr == searchChar) //use if that check *ptr value is equal to searchChar

       {

           searchResult = ptr;//holding ptr value in searchResult

       }

       ptr++;//incrementing ptr value

   }

   if (searchResult != nullptr) //use if that checks searchResult value not equal to nullptr

   {

       cout << "Character found." << endl;//print message

   }  

   else //else block

   {

       cout << "Character not found." << endl;//print message

   }

   return 0;

}  

Output:

Character found.

Explanation:

In this code inside the main method, a char array is declared that holds values.  

In the next step, some char variable and pointer type char variable are declared that holds values.

Inside the while loop, the pointer variable is used, and in the if the block it checks its value and holds its values.

Outside the loop, it checks searchResult value not equal to nullptr and prints the value as per the given condition.

Answer:

deez lOL LOLOLOLOL

Explanation:

Answer:

DVD-R is a good choice for long-term archiving because the stored data do not degrade over time. This is due to its write-once, read-many nature, which makes it less likely to degrade compared to magnetic tape, HDDs or solid-state drives (SSDs).

Explanation:

Answer: A) File

Just took it (:

Answer:

a-file

Explanation:

its easy i just took the assignment