a. //Initialize the variables int n=1; int k=2; int r=n; //Check the condition if (k<n) { //Assign k to r r=k } Value of each variable after “ if ” statement: n=1, k=2, r=1 Explanation: The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Question

Want to see more full solutions like this?

Answer 1

Expert Solution & Answer

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

using r language for integration theta = integral 0 to infinity (x^4)*e^(-x^2)/2 dx (1) use the density function of standard normal distribution N(0,1) f(x) = 1/sqrt(2pi) * e^(-x^2)/2 -infinity <x<infinity as importance function and obtain an estimate theta 1 for theta set m=100 for the estimate whatt is the estimate theta 1? (2)use the density function of gamma (r=5 λ=1/2)distribution f(x)=λ^r/Γ(r) x^(r-1)e^(-λx) x>=0 as importance function and obtain an estimate theta 2 for theta set m=1000 fir the estimate what is the estimate theta2? (3) use simulation (repeat 1000 times) to estimate the variance of the estimates theta1 and theta 2 which one has smaller variance?

using r language A continuous random variable X has density function f(x)=1/56(3x^2+4x^3+5x^4).0<=x<=2 (1) secify the density g of the random variable Y you find for the acceptance rejection method. (2) what is the value of c you choose to use for the acceptance rejection method (3) use the acceptance rejection method to generate a random sample of size 1000 from the distribution of X .graph the density histogram of the sample and compare it with the density function f(x)

using r language a continuous random variable X has density function f(x)=1/4x^3e^-(pi/2)^4,x>=0 derive the probability inverse transformation F^(-1)x where F(x) is the cdf of the random variable X

Answer 2

Expert Solution & Answer

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Expert Solution & Answer

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Expert Solution & Answer

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Answer 8

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Answer 9

Chapter 5, Problem 1RE

Explanation of Solution

a.

//Initialize the variables

int n=1; int k=2; int r=n;

//Check the condition

if (k<n)

{

//Assign k to r

r=k

}

Value of each variable after “if” statement:

n=1, k=2, r=1

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “n” is not greater than “k” and so the value of “r” remains unchanged.

Answer 10

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Answer 11

Explanation of Solution

b.

//Initialize the variables

int n=1; int k=2; int r;

//Check the condition

if (n<k)

{

    //Assign k to r

    r=k

}

//Otherwise

else

{

    //Add k+n and store it in r

    r=k+n;

}

Value of each variable after “if”statement:

n=1, k=2, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement is true. As the value of “k” is greater than “n” and so the value of “r” is changed to “2”.

Answer 12

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Answer 13

Explanation of Solution

c.

//Initialize the variables

int n=1; int k=2; int r=k;

//Check the condition

if (r<k)

{

    //Assign k to r

    n=r

}

//Otherwise

else

{

    //Assign n to k

    k=n;

}

Value of each variable after “if”statement:

n=1, k=1, r=2

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “k” and it moves to the “else” part where the value of “k” is changed to value of “n” that is “1”. In the first statement, the value of “r” is initialized to value of “k” that is “2”.

Answer 14

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Answer 15

Explanation of Solution

d.

//Initialize the variables

int n=1; int k=2; int r=3;

//Check the condition

if (r<n+k)

{

    //Assign k to r

    r=2n

}

//Otherwise

else

{

    //Assign n to k

    k=2r;

}

Value of each variable after “if”statement:

n=1, k=6, r=3

Explanation:

The value assigned to variable “n” is “1”, the condition inside the “if” statement fails. As the value of “r” is not greater than “n+k” and it moves to the “else” part where the value of “k” is changed to value of “2*r” that is “6”. In the first statement, the value of “r” is initialized to “3”.

Answer 16

Ch. 5.1 - In some Asian countries, the number 14 is...Ch. 5.1 - Consider the following if statement to compute a...Ch. 5.1 - Prob. 3SC Ch. 5.1 - Prob. 4SC Ch. 5.1 - The variables fuelAmount and fuelCapacity hold the...Ch. 5.2 - Prob. 6SC Ch. 5.2 - Prob. 7SC Ch. 5.2 - Prob. 8SC Ch. 5.2 - Prob. 9SC Ch. 5.2 - Prob. 10SC

Videos

Explanation of Solution

Explanation of Solution

Explanation of Solution

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions