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The marginal product of labour function for International Trading Inc. is given by the equation
MPL = 10 K0.5/ L0.5
Currently, the firm is using units of capital and 121 units of labour. Given the very specialized nature of the capital equipment, it takes six to nine months to increase the capital stock, but the rate of labour input can be varied daily. If the price of labour is $10 per unit and the price of output is $2 per unit, is the firm operating efficiently in the short run? If not, explain why, and determine the optimal rate of labour input.
The crane is pulling up with a force of 216N and the box is being down by gravity with a force of 127 N draw a free body diagram and determine what the net force is and in what direction it is acting.
2. A long straight wire has a charge per unit length of 3.00 x 10^-12 C/m. At what distance from the wire is the electric field equal to 0.600 N/c.
Derrive marginal value product and average value product equation using the production function Y=2X+16X2-0.4X3
- Use the production function Q = 10K0.5L0.6 to complete the following production table.
Rate of Capital Input (K)
6 24.5 56.3 71.8
5
4 30.3
3 45.5
2 27.3
1 10.1 29.9
1 2 3 4 5 6
Rate of Labour input (L)
a. For this production system, are returns to scale decreasing, constant, or increasing?
b. Suppose the wage rate is $28, the price of capital also is $28 per unit, and the firm currently is producing 30.3 units of output per period using four units of capital and two units of labour. Is this an efficient resource combination? Why? ( Hint: Compare the marginal products of capital and labour at initial combination.)
import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner in = new Scanner(System.in); int house = 5; int room = 3; int len = 2; int we = 4; double cost = 6; int areaRoom = len * we; int totalArea = areaRoom * room * house; double totalCost = totalArea * cost; System.out.println("Room area: " + areaRoom); System.out.println("Area of all rooms: " + totalArea); System.out.println("Total price of the construction: " + totalCost); } }
I want pseudocode and flowchart of this code.
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int[] arr = new int[7]; boolean isSorted = true; int buf; for (int i = 0; i < arr.length; i++) { System.out.print("Month " + (i + 1) + ": "); arr[i] = in.nextInt(); } for (int i = 0; i < arr.length; i++){ for (int j = 0; j < arr.length - 1; j++) { if (arr[i] < arr[j]) { buf = arr[i]; arr[i] = arr[j]; arr[j] = buf; } } } for (int i : arr) { System.out.print(i + " "); } } }
This is the code i want pseudocode and flowchart of this code.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int[] arr = new int[9];
int[] rez = new int[9];
for (int i = 0; i < arr.length; i++) {
System.out.print("Car " + (i + 1) + ": ");
arr[i] = in.nextInt();
}
int x = 0;
for (int i : arr) {
if(i%2 == 0){
rez[x] = i;
x++;
}
}
for (int i : arr) {
if(i%2 != 0){
rez[x] = i;
x++;
}
}
for (int i : rez) {
System.out.print(i + " ");
}
}
}
This is the code i want pseudocode and flowchart of this code.
Consider the abstract class declaration:
class aPolygon
{
protected:
long double * sides, // There are N sides with different lengths
* angles; // There are N angles with different sizes
long int no_of_sides; // The number of sides, N, of this polygon
public: aPolygon(long int _sides = 1);
/* It sets no_of_sides = _sides.*/
virtual void input () = 0; // input the derived polygon
virtual void display () = 0; // display the derived polygon
~aPolygon() {}
};
Write a complete C++ class implementation of the following derived class, aSquare, with the class interface::
class aSquare : public aPolygon
{
public:
aSquare(); /* It creates a square as a unit square */
~aSquare() {}
virtual void input(); /* It inputs the data for the square */
virtual void display(); /* It displays the square: angles, sides, perimeter and area */
long double perimeter(); /* It computes the perimeter of the square */
long double area(); /* It computes the area of the square: */
};
A sample of 89.1 g of tetraphosphorous decoxide (P4O10) reacts with 49.0 g of water to produce phosphoric acid (H3PO4) according to the following balanced equation.
P4O10+6H2O⟶4H3PO4
Determine the limiting reactant for the reaction.
Calculate the mass of H3PO4 produced in the reaction.
Calculate the percent yield of H3PO4 is isolated after carrying out the reaction.
