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Zielflug [23.3K]
3 years ago
8

What is a positive and negative interface of Apple, Microsoft, Linux GUI

Computers and Technology
2 answers:
Gala2k [10]3 years ago
5 0
The other guy said what I was gonna say lol his is correct
Vsevolod [243]3 years ago
3 0

Answer:

Microsoft is perfect the only negative is their tech support

Apple is AMAZING but their software is confusing and doesn't let you customize

Linux GUI its good but its way behind in its time

Explanation:

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This is the formula for the future worth of an investment over time with consistent additional monthly payments and a steady rat
stepladder [879]

Answer:

The code is given in C++ below

Explanation:

#include <iostream>

using namespace std;

int main()

{

float fv,pv,r,k,n,pmt,totalmoneyinvested;

pv=1000.00;

r=6/100;

k=12; //The value of k should be 12 for monthly installments

n=45;

pmt=250;

totalmoneyinvested=pv+(pmt*12*45); //The total money you invested

 

fv=pv*(1+r/k)*n*k+pmt*((1+r/k)*n*k-1)*(1+r/k)*r/k;

 

cout<<"Initial Investment:"<<" $"<<pv;

cout<<"\nRate Of Return:6%";

cout<<"\nLength of Time:"<<n<<"year";

cout<<"\nMonthly Payment:"<<" $"<<pmt;

cout<<"\nFinal Amount:"<<" $"<<fv;

cout<<"\nThe Money You Invested Is $"<<totalmoneyinvested<<" And The Final Amount Is $"<<fv;

return 0;

}

4 0
3 years ago
Ok so another weird question! So if you know what google drive is and how to upload a video why does it keep adding hours! it ke
lesya [120]

Answer:Your video may be too long and you may not have that much storage left.

Explanation:

5 0
3 years ago
Concept tests in the screening and evaluation stage of the new-product process rely on written descriptions, sketches, or mock-u
pochemuha

Answer:

The answer is "Actual products".

Explanation:

The key purpose of this research is to better analyze the concepts, which is done to establish customer buying expectations and perceptions of the item. The key point is to assess the consumers ' initial response to the product idea.

  • This concept testing is also known as a creation, it is an advantage, which can be conveyed to the user to test their reactions.  
  • In conceptual testing, it is a quality check between the design definition and the actual production of the product.

6 0
3 years ago
given that play_list has been defined to be a list, write an expression that evaluates to a new list containing the elements at
Tamiku [17]

Answer:

new_list = play_list[0:4]

Explanation:

5 0
3 years ago
Write a C++ program that reads from the standard input and counts the number of times each word is seen. A word is a number of n
Rufina [12.5K]

Answer:

#include <stdio.h>

#include <string.h>

#include <ctype.h>

 

struct detail

{

   char word[20];

   int freq;

};

 

int update(struct detail [], const char [], int);

 

int main()

{

   struct detail s[10];

   char string[100], unit[20], c;

   int i = 0, freq = 0, j = 0, count = 0, num = 0;

 

   for (i = 0; i < 10; i++)

   {

      s[i].freq = 0;

   }

   printf("Enter string: ");

   i = 0;

   do

   {

       fflush(stdin);

       c = getchar();

       string[i++] = c;

 

   } while (c != '\n');

   string[i - 1] = '\0';

   printf("The string entered is: %s\n", string);

   for (i = 0; i < strlen(string); i++)

   {

       while (i < strlen(string) && string[i] != ' ' && isalnum(string[i]))

       {

           unit[j++] = string[i++];

       }

       if (j != 0)

       {

           unit[j] = '\0';

           count = update(s, unit, count);

           j = 0;

       }

   }

 

   printf("*****************\nWord\tFrequency\n*****************\n");

   for (i = 0; i < count; i++)

   {

       printf("%s\t   %d\n", s[i].word, s[i].freq);

       if (s[i].freq > 1)

       {

           num++;

       }

   }

   printf("The number of repeated words are %d.\n", num);

 

   return 0;

}

 

int update(struct detail s[], const char unit[], int count)

{

   int i;

 

   for (i = 0; i < count; i++)

   {

       if (strcmp(s[i].word, unit) == 0)

       {

           s[i].freq++;

 

           return count;

       }

   }

   /*If control reaches here, it means no match found in struct*/

   strcpy(s[count].word, unit);

   s[count].freq++;

 

   /*count represents the number of fields updated in array s*/

   return (count + 1);

}

4 0
3 years ago
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