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Stolb23 [73]
3 years ago
11

#include

Computers and Technology
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

const int NUM_VALS = 4;

vector<int> testGrades(NUM_VALS);

int i = 0;

int sumExtra = -9999; // Assign sumExtra with 0 before your for loop

testGrades.at(0) = 101;

testGrades.at(1) = 83;

testGrades.at(2) = 107;

testGrades.at(3) = 90;

/* Your solution goes here */

sumExtra = 0;

for(i = 0; i <= testGrades.size() -1; i++){

   if(testGrades.at(i) > 100){

       sumExtra = sumExtra + (testGrades.at(i) - 100);

   }

}

cout << "sumExtra: " << sumExtra << endl;

return 0;

}

Explanation:

Looks like you almost solved the question. I highlighted the parts you have been missing above.

In order to use vectors in C++, you need to add the vector library at the beginning of the program, #include <vector>

In order to initialize the vector, you need to specify its type inside <int>, since we work with the integers in the question the type must be <em>int</em>

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In 3–5 sentences, describe how technology helps business professionals to be more efficient.
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Answer:

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Explanation:

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Which of the following is usually used to connect a monitor to a computer?
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DVI cable

Explanation:

DVI is for video signals.

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Write a C program to prform simple C aritlimetic calculations. The user is to enter a simple expression(integer operaior integer
larisa86 [58]

Answer:

Here is the C program:

#include <stdio.h>  //to use input output functions

//functions prototype

unsigned int mod(unsigned int a, unsigned int b);  

unsigned int mul(unsigned int a, unsigned int b);  

unsigned int sub( unsigned int a,unsigned int b);

float divide(unsigned int a,unsigned int b);  

unsigned int add( unsigned int a,unsigned int b);  

int main()  {  //start of main method

unsigned int a, b;   //declare variables to store the operands

char d;  //declare variable to store the operator

printf("Enter an operator:  ");   //prompts user to enter an operator

scanf("%c",&d);  //reads the operator from use

getchar();  //gets a character

while (d!='q')   { //keeps iterating until user enters q to quit

printf("Enter 1st operand: ");   //prompts user to enter first operand

scanf("%d",&a);   //reads first operand from user

getchar();  //reads character

printf("Enter 2nd operand: ");   //prompts user to enter second operand

scanf("%d",&b);   //reads second operand from user

getchar();  

if (d=='%')  {   //if the character of operator is a mod

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

mod(a,b);  }  //displays computed modulo of two input operands

if (d=='*')   //if the input character is for multiplication operator

{printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

mul(a,b); }  //displays computed multiplication

if (d=='+')  {  //if the input character is for addition operator

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = "); // displays =

add(a,b);  }   //displays computed addition

if (d=='/')  {  //if the input character is for division operator

printf("%d",a); // prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

divide(a,b);  }   //displays computed division

if (d=='-')  {  //if the input character is for subtraction operator

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b); // displays operand 2

printf(" = ");  //displays =

sub(a,b);  }  //displays computed subtraction

printf("Enter an operator: ");   //asks again to enter an operator

scanf("%c",&d);  //reads operator from user

getchar();  }  }   //gets character

unsigned int mod( unsigned int a, unsigned int b){  //function to compute modulo of two integers with no sign

     int c = a%b;  //computes mod

    printf("%d",c);  }  //displays mod result

unsigned int add(unsigned int a, unsigned int b){     // function to compute addition of two integers

    int c = a+b; //computes addition

    printf("%d\n",c);  } //displays result of addition

unsigned int mul(unsigned int a, unsigned int b){       //function to compute multiplication of two integers

    int c = a*b;  //multiplies two integers

   printf("%d\n",c); }  //displays result of multiplication

float divide(unsigned int a, unsigned int b){   //function to compute division of two integers

    float c = a/b;  //divides two integers and stores result in floating point variable c

    printf("%f\n",c);  } //displays result of division

unsigned int sub(unsigned int a, unsigned int b){       //function to compute subtraction of two integers

    int c = a-b;  //subtracts two integers

    printf("%d\n",c);  }  //displays result of subtraction

Explanation:

The program is well explained in the comments mentioned with each line of the program. The program uses while loop that keeps asking user to select an operator and two operands to perform arithmetic calculations. The if conditions are used to check what arithmetic calculation is to be performed based on user choice of operand and calls the relevant function to perform calculation. There are five functions that perform addition, modulo, subtraction, division and multiplication and display the result of computation. The screenshot of output is attached.

4 0
3 years ago
What is a router?
DIA [1.3K]

Answer:

A) a device that sends data to the receiving device

Explanation:

3 0
3 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
3 years ago
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