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Natali [406]
3 years ago
12

Which of the following methodologies might be most appropriate if you have a system project with:unclear user requirements; unfa

miliar technologies; somewhat complex; needs to be reliable; time isnot an issue and the schedule visibility is somewhat important?
a) Waterfall
b) Parallel
c) Iterative
d) System prototyping
e) Throwaway prototyping
Computers and Technology
1 answer:
andreyandreev [35.5K]3 years ago
7 0

Answer: Throwaway prototyping

Explanation:

Waterfall is used when the system has clear requirements, reasonably reliable, very familiar technologies, isn't too complex. It's also used when there's a very long time schedule and in a scenario whereby the schedule visibility isn't important.

Parallel is used when the system project has clear requirements, shirt time schedule, technologies that are very familiar, not all that complex; reasonably reliable, and the schedule visibility is not important.

With regards to the question, the description given fits a throwaway prototyping. Therefore, the correct option is E.

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Why does binary addition have a rule for 1 + 1 + 1 if only two numbers<br> are being added together?
dalvyx [7]

Answer:

See below

Explanation:

Because sometimes you have to 'carry' a 1 over to the ext column when adding two binary numbers

Example :

 1 1 1

<u>+1 1 1 </u>    <==== starting in the first R column add  1 + 1  to get 0 and carry 1

                then the next column you will add   1 + 1 + 1  = 1   and carry 1 again

                    then 1 + 1 + 1= 1  and carry 1 again (to column 4)  to get

1   1  1 0

5 0
2 years ago
Explain demand paging with a proper example
julia-pushkina [17]
<h2><em><u>Demand</u></em><em><u> </u></em><em><u>paging</u></em><em><u> </u></em><em><u>:</u></em></h2>

<h3><em>In computer operating systems, demand paging (as opposed to anticipatory paging) is a method of virtual memory management. In a system that uses demand paging, the operating system copies a disk page into physical memory only if an attempt is made to access it and that page is not already in memory (i.e., if a page fault occurs). It follows that a process begins execution with none of its pages in physical memory, and many page faults will occur until most of a process's working set of pages are located in physical memory. This is an example of a lazy loading technique.</em></h3>

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6 0
3 years ago
Write a function which the counts the number of odd numbers and even numbers currently in the stack and prints the results.
Lerok [7]

Answer:

See the code below and the algorithm explanation on the figure.

Explanation:

The explanation in order to get the answer is given on the figure below.

Solving this problem with C. The program is given below:

#include <stdio.h>

int main(void) {

   int n, Even=0, Odd=0, Zeros=0;  

   for (;;) {

       printf("\nEnter the value the value that you want to check(remember just integers): ");

       //IF we input a non-numeric character the code end;

       if (scanf("%d", &n) != 1) break;

       if (n == 0) {

           Zeros++;

       }

       else {

           if (n % 2) {

               Even++;

           }

           else {

               Odd++;

           }

       }

   }  

   printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);

   return 0;

}

5 0
3 years ago
Write a program that prompts the user to enter in a postive number. Only accept positive numbers - if the user supplies a negati
marusya05 [52]

Answer:

The c++ program to check prime numbers is shown below.

#include <iostream>

using namespace std;

int main() {

   int num, prime=0;

   do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

   

   if(num==1 || num==2 || num==3)

       cout<<num<<" is a prime number."<<endl;

   else if(num%2 == 0)

       cout<<num<<" is not a prime number."<<endl;

   else

   {

       for(int k=3; k<num/2; k++)

       {

           if(num%k == 0)

               prime++;

       }    

   if(prime>1)

       cout<<num<<" is not a prime number."<<endl;

   else

       cout<<num<<" is a prime number."<<endl;

   }

}

OUTPUT

Enter a positive number.

-7

Invalid number. Enter a positive number

0

Enter a positive number.

79

79 is a prime number.

Explanation:

The user input is validated for positivity. A do while loop along with an if statement is implemented for verification.

do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

The test for prime number is done by using multiple if else statements.

If user inputs 1, 2, or 3, message is displayed.

Else If user inputs an even number, message is displayed for not prime. This is done by taking modulo of the number upon division by 2.

Else if user inputs neither an even number nor a number less than 3, the modulus of the number is taken with divisors beginning from 3 up to half of the input number.

For this, an integer variable prime is initialized to 0. A number can be completely divisible by itself or by its factors.

If the number is divisible by any of the divisors, value of variable prime is increased by 1. If value of prime is greater than 1, this means that the user input is divisible by more than one divisor. Hence, the given number is not a prime number.

5 0
3 years ago
Legal counsel has notified the information security manager of a legal matter that will require the preservation of electronic r
Ray Of Light [21]

Answer: (B) Legal hold

Explanation:

As being a security director, the security director are approached to solidify a server that run the red hat enterprise server the 5.5.

The servers are being utilized as DNS (Dynamic host service) and the server time. This is not utilized as the database, print server and the web server. There is no particular remote associations with the server.

The window of the command will be given the root accessing. Then, it provide the associated by means of a protected shell with the access root in the system.

 

3 0
3 years ago
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