Answer:
a.both the equilibrium price and quantity of DVD players will decrease
Explanation:
When the amount required or given varies, even if the price stays the same, a move in the demand or supply curve occurs. Changes in the curve of demand mean that the initial relationship of production has shifted so that demand of quantity has a factor apart from cost influenced
A right shift change of the supply curve shows an increase in supply and, on equal footing, the equilibrium price decreases.
Once the demand curve shifts to the left, the demand decreases.
Answer:
The program in C++ is as follows:
#include <fstream>
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
string filename;
cout<<"Filename: ";
cin>>filename;
ifstream inFile(filename);
if(!inFile) {
cout << endl << "Cannot open file " << filename;
return 1; }
ofstream fout;
ifstream fin;
fin.open("invalid-numbers.txt");
fout.open ("invalid-numbers.txt",ios::app);
double sum = 0; int valid = 0; int invalid = 0;
double num = 0;
while(!inFile.eof()) {
inFile >> num;
if(num >= 0 && num<=110){ sum+=num; valid++; }
else{ invalid++;
if(fin.is_open()){
fout<<fixed<<setprecision(2)<<num<<"\n"; } } }
fin.close();
fout.close();
inFile.close();
cout<<"Total values: "<<valid+invalid<<endl;
cout<<"Invalid values: "<<invalid<<endl;
cout<<"Valid values: "<<valid<<endl;
cout<<"Average of valid values: "<<fixed<<setprecision(2)<<sum/valid<<endl;
double inv;
ifstream inFiles("invalid-numbers.txt");
while(!inFiles.eof()) {
inFiles >> inv;
cout<<inv<<"\n";
}
inFiles.close();
return 0;
}
Explanation:
See attachment for source file where comments are used to explain each line
Ill choose flowchart. Look picture for the answer
ask me if you have any doubts about my answer.
Answer:
The program in Python is as follows:
num1 = int(input())
num2 = int(input())
if num1 >=0 and num2 >= 0:
print(num1+num2)
elif num1 <0 and num2 < 0:
print(num1*num2)
else:
if num1>=0:
print(num1**2)
else:
print(num2**2)
Explanation:
This gets input for both numbers
num1 = int(input())
num2 = int(input())
If both are positive, the sum is calculated and printed
<em>if num1 >=0 and num2 >= 0:</em>
<em> print(num1+num2)</em>
If both are negative, the products is calculated and printed
<em>elif num1 <0 and num2 < 0:</em>
<em> print(num1*num2)</em>
If only one of them is positive
else:
Calculate and print the square of num1 if positive
<em> if num1>=0:</em>
<em> print(num1**2)</em>
Calculate and print the square of num2 if positive
<em> else:</em>
<em> print(num2**2)</em>
