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liraira [26]
3 years ago
10

C++

Computers and Technology
2 answers:
egoroff_w [7]3 years ago
5 0

Answer:

def driving_cost(driven_miles, miles_per_gallon, dollars_per_gallon):

  gallon_used = driven_miles / miles_per_gallon

  cost = gallon_used * dollars_per_gallon  

  return cost  

miles_per_gallon = float(input(""))

dollars_per_gallon = float(input(""))

cost1 = driving_cost(10, miles_per_gallon, dollars_per_gallon)

cost2 = driving_cost(50, miles_per_gallon, dollars_per_gallon)

cost3 = driving_cost(400, miles_per_gallon, dollars_per_gallon)

print("%.2f" % cost1)

print("%.2f" % cost2)

print("%.2f" % cost3)

Explanation:

Cerrena [4.2K]3 years ago
4 0

Answer:

#include <iostream>

#include<iomanip>

using namespace std;

double DrivingCost(double drivenMiles, double milesPerGallon, double dollarsPerGallon)

{

  double dollarCost = 0;

  dollarCost = (dollarsPerGallon * drivenMiles) / milesPerGallon;

  return dollarCost;

}

int main()

{

  double miles = 0;

  double dollars = 0;

  cout << "Enter miles per Gallon   : ";

  cin >> miles;

  cout << "Enter dollars per Gallon: ";

  cin >> dollars;

  cout << fixed << setprecision(2);

  cout << endl;

  cout << "Gas cost for 10 miles : " << DrivingCost(10, miles, dollars) << endl;

  cout << "Gas cost for 50 miles : " <<DrivingCost(50, miles, dollars) << endl;

  cout << "Gas cost for 400 miles: "<<DrivingCost(400, miles, dollars) << endl;

  return 0;

}

Explanation:

  • Create a method definition of DrivingCost that accepts  three input double data type parameters drivenMiles,  milesPerGallon, and dollarsPerGallon and returns  the dollar cost to drive those miles .
  • Calculate total dollar cost and store in the variable, dollarCost .
  • Prompt and read the miles and dollars per gallon  as input from the user .
  • Call the DrivingCost function three times  for the output to the gas cost for 10 miles,  50 miles, and 400 miles.

 

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Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
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Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)    

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

Explanation:  

Unification: Any substitution that makes two expressions equal is called a unifier.  

a) P(B,A,B), P(x,y,z)  

Use { x/B}  

=> P(B,A,B) , P(B,y,z)  

Now use {y/A}  

=> P(B,A,B) , P(B,A,z)  

Now, use {z/B}  

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }  

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q  

c. Older(Father(y),y), Older(Father(x),John)  

Use {y/x}  

=> Older(Father(x),x), Older(Father(x),John)  

Now use { x/John }  

=> Older(Father(John), John), Older(Father(John), John)  

Thus , mgu ={ y/x , x/John }  

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))  

Use { y/x }  

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}  

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B  

e) P(f(x), x, g(x)), P(f(y), A, z)  

Use {x/y}  

=> P(f(y), y, g(y)), P(f(y), A, z)  

Now use {z/g(y)}  

P(f(y), y, g(y)), P(f(y), A, g(y))  

Now use {y/A}  

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

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Answer:

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Both conditions are false

Hence, false values are returned.

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vekshin1

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Answer:

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