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julia-pushkina [17]
3 years ago
13

What is the opportunity cost of computers when moving from point A to point B? -15 DVDs b. What is the opportunity cost of compu

ters when moving from point B to point C? -18 DVDs c. What is the opportunity cost of computers when moving from point C to point D? -20 DVDs d. What is the opportunity cost of computers when moving from point D to point E? DVDs e. As we produce more computers, opportunity costs are .
Computers and Technology
1 answer:
oee [108]3 years ago
5 0

Answer:

Increasing

Explanation:

Generally in economics, opportunity cost is the benefit that a person, a business, or an investor forgo or missed when he chooses one alternative instead of the other.

Under production, opportunity cost occurs when a producer sacrifice the production of one for the production of another good. Using good X and Y as examples, opportunity cost is measured by the number of units of good Y that the producer gives up in order to produce one or more units of good X.

In the question, the two products used are computers and DVD. The -15, -18 and -20 implies that producing an extra unit of computer by moving from point A to point B, point B to C, and point C to D, 15, 18 and 20 units of DVD respectively have to be given up.

Since 20 is greater than 18 and 18 is also greater than 15, it therefore implies as we produce more computers, the opportunity cost, which are the number of DVDs given up at each point, are increasing.

I wish you the best.

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Answer:

See explaination

Explanation:

program code.

/* PRE PROCESSOR DIRECTIVES */

#include<stdio.h>

/* PRE-DEFINED VALUES FOR TMAX AND TMIN */

#define TMax 2147483647

#define TMin (-TMax -1)

/* saturating_add(int,int) METHOD IS CALLED HERE */

int saturating_add(int firstNumber, int secondNumber)

{

/*

FOR BETTER UNDERSTANDING, LETS TAKE TEST CASE,

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int w = sizeof(firstNumber) << 3;

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*/

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/*

MASK INTEGER VARIABLE IS TAKEN

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*/

int mask = 1 << (w - 1);

/* FIRST NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbFirstNumber = firstNumber & mask;

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/* MAIN FUNCTION STARTS HERE */

int main(){

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/* DISPLAY THE RESULT OF TEST CASE */

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Answer:

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