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Andreas93 [3]
2 years ago
13

Write pseudocode for washing a car using at least five steps.

Computers and Technology
1 answer:
Travka [436]2 years ago
5 0

Answer:

#include <iostream>

int main()

{

bool carWashed{ 0 };

bool washCar{ 0 };

while(carWashed == 0)

{

 washCar();

}

if(washCar == 1)

{

carWashed{ 1 }

}

}

Explanation:

c++

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Who is the king of computers?
anygoal [31]

Answer: <u>Bill Gate</u>, who is known as the king of computer programs

Hope this helps!

4 0
2 years ago
Read 2 more answers
Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
Evgen [1.6K]

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 }.

7 0
3 years ago
In order to consolidate your theoretical knowledge into technique and skills with practical and applicational value, you will us
Arisa [49]

Answer:

Check the explanation

Explanation:

Lasso: R example

To run Lasso Regression you can re-use the glmnet() function, but with the alpha parameter set to 1.

# Perform 10-fold cross-validation to select lambda --------------------------- lambdas_to_try <- 10^seq(-3, 5, length.out = 100) # Setting alpha = 1 implements lasso regression lasso_cv <- cv.glmnet(X, y, alpha = 1, lambda = lambdas_to_try, standardize = TRUE, nfolds = 10) # Plot cross-validation results plot(lasso_cv)

Best cross-validated lambda lambda_cv <- lasso_cv$lambda.min # Fit final model, get its sum of squared residuals and multiple R-squared model_cv <- glmnet(X, y, alpha = 1, lambda = lambda_cv, standardize = TRUE) y_hat_cv <- predict(model_cv, X) ssr_cv <- t(y - y_hat_cv) %*% (y - y_hat_cv) rsq_lasso_cv <- cor(y, y_hat_cv)^2 # See how increasing lambda shrinks the coefficients -------------------------- # Each line shows coefficients for one variables, for different lambdas. # The higher the lambda, the more the coefficients are shrinked towards zero. res <- glmnet(X, y, alpha = 1, lambda = lambdas_to_try, standardize = FALSE) plot(res, xvar = "lambda") legend("bottomright", lwd = 1, col = 1:6, legend = colnames(X), cex = .7)

Kindly check the Image below.

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3 years ago
Which of these is system software? Check all
Darya [45]

Answer:

Windows OS is software....

6 0
3 years ago
Enzymes_____________.
lord [1]

Answer:

b. speed up chemical reactions.

Explanation:

brainly plz

4 0
2 years ago
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