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Grace [21]
3 years ago
7

When you must give something up in order to get something else it's called

Computers and Technology
1 answer:
matrenka [14]3 years ago
8 0

Answer:

Opportunity cost refers to what you have to give up to buy what you want in terms of other goods or services. When economists use the word “cost,” we usually mean opportunity cost.

Explanation:

please give me a heart

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scientist think that two species with homologous structures probably had ____ A. Ancestor B. Fossil C. Environment D. Offspring
Naddika [18.5K]
The answer is A: ancestor
4 0
3 years ago
Graduating from college can boost your income 60% compared to high school graduates. An average high school graduate without a c
Aleksandr-060686 [28]

Answer: $2,240,000

Explanation:

If you multiply 35,000 by 1.6, you will get 56,000.

Then you multiply 56,000 by 40 and you will get $2,240,000.

8 0
3 years ago
The following section of psudo code inputs a number n, multiplies together 1 x 2 x 3 x ..... x n, calculates input number/sum an
xenn [34]

Answer:

INPUT n

sum = 1

FOR mult = 1 TO n

sum = sum*mult

NEXT

result = n/sum

PRINT result

Explanation:

sum should not be defined inside the for loop because it will reset to the initial number after every loop iteration which is not what is desired. Sum should not be 0 if it is being multiplied then divided. This will cause an error. Result should be calculated after sum has settled on a final number, after the loop has finished.

4 0
3 years ago
1-(50 points) The function sum_n_avgcomputes the sum and the average of three input arguments and relays its results through two
MAXImum [283]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

double *sum_n_avg(double n1, double n2, double n3);

int main () {

   double n1, n2, n3;

   cout<<"Enter 3 inputs: ";

   cin>>n1>>n2>>n3;

   double *p;

   p = sum_n_avg(n1,n2,n3);

   cout<<"Sum: "<<*(p + 0) << endl;

   cout<<"Average: "<<*(p + 1) << endl;

  return 0;

}

double *sum_n_avg(double n1, double n2, double n3) {

  static double arr[2];

  arr[0] = n1 + n2 + n3;

  arr[1] = arr[0]/3;

  return arr;

}

Explanation:

This defines the function prototype

double *sum_n_avg(double n1, double n2, double n3);

The main begins here

int main () {

This declares variables for input

   double n1, n2, n3;

This prompts the user for 3 inputs

   cout<<"Enter 3 inputs: ";

This gets user inputs

   cin>>n1>>n2>>n3;

This declares a pointer to get the returned values from the function

   double *p;

This passes n1, n2 and n3 to the function and gets the sum and average, in return.

   p = sum_n_avg(n1,n2,n3);

Print sum and average

<em>    cout<<"Sum: "<<*(p + 0) << endl;</em>

<em>    cout<<"Average: "<<*(p + 1) << endl;</em>

  return 0;

}

The function begins here

double *sum_n_avg(double n1, double n2, double n3) {

Declare a static array

  static double arr[2];

Calculate sum

  arr[0] = n1 + n2 + n3;

Calculate average

  arr[1] = arr[0]/3;

Return the array to main

  return arr;

}

5 0
3 years ago
A favorite pastime of information security professionals is ____, which is a simulation of attack and defense activities using r
GenaCL600 [577]

Answer:

war gaming

Explanation:

Wargaming in hacking is a cyber security challenge where competitors must exploit or defend a loophole/vulnerability in a system or application or gain or prevent access to a particular computer system, it is in a form of competition but it really helps them in securing systems/application when it arises.

5 0
4 years ago
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