Physics: 4. Find the value of W2 if W1 = 250N.

4. Find the value of W2 if W1 = 250N., №10595243, 27.04.2022 04:54

Computers and Technology

This question involves the implementation of the PasswordGenerator class, which generates strings containing initial passwords for online accounts. The PasswordGenerator class supports the following functions: - Creating a password consisting of a specified prefix, a period, and a randomly generated numeric portion of specified length - Creating a password consisting of the default prefix A , a period, and a randomly generated numeric portion of specified length - Reporting how many passwords have been generated The following table contains a sample

Step-by-step answer

P Answered by PhD

40

The following code will be used for the PasswordGenerator class.

Explanation:

import java.util.Random;

public class PasswordGenerator {

private static int passwordsGenerated =0;

private static Random random = new Random();

private String prefix;

private int length;

public PasswordGenerator(int length,String prefix) {

this.prefix = prefix;

this.length = length;

}

public PasswordGenerator(int length) {

this.prefix = "A";

this.length = length;

}

public String pwGen(){

String pwd= this.prefix+".";

for(int i=1;i<=this.length;i++){

pwd+=random.nextInt(10);

}

passwordsGenerated+=1;

return pwd;

}

public int pwCount(){

return passwordsGenerated;

}

public static void main(String[] args) {

PasswordGenerator pw1 = new PasswordGenerator(4,"chs");

System.out.println(pw1.pwCount());

System.out.println(pw1.pwGen());

System.out.println(pw1.pwCount());

PasswordGenerator pw2 = new PasswordGenerator(6);

System.out.println(pw2.pwCount());

System.out.println(pw2.pwGen());

System.out.println(pw2.pwCount());

System.out.println(pw1.pwCount());

}

Mathematics

Suppose that W1 is a random variable with mean μ and variance σ21 and W2 is a random variable with mean μ and variance σ2. From Example 5.4.3, we know that cW1 + (1 − c)W2 is an unbiased estimator of μ for any constant c 0. If W1 and W2 are independent, for what value of c is the estimator cW1 + (1 − c)W2 most efficient?

Step-by-step answer

P Answered by Specialist

Step-by-step explanation:

The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.

The detailed analysis and step by step approach is as shown in the attachment.

Mathematics

Suppose that W1 is a random variable with mean μ and variance σ21 and W2 is a random variable with mean μ and variance σ2. From Example 5.4.3, we know that cW1 + (1 − c)W2 is an unbiased estimator of μ for any constant c 0. If W1 and W2 are independent, for what value of c is the estimator cW1 + (1 − c)W2 most efficient?

Step-by-step answer

P Answered by Specialist

Step-by-step explanation:

The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.

The detailed analysis and step by step approach is as shown in the attachment.

Computers and Technology

Univariate linear regression Note: Solutions to this problem must follow the method described in class and the linear regression handout. There is some flexibility in how your solution is coded, but you may not use special functions that automatically perform linear regression for you. Load in the BodyBrain Weight.csv dataset. Perform linear regression using two different models: M1: brain_weight = w0 + w1 x body_weight M2: brain_weight = w0 + w1 x body_weight + w2 x body_weight2 a. For each model, follow the steps shown in class to solve for w.

Step-by-step answer

P Answered by Specialist

regression line is Y = 124.9281 + 0.9370 *x

SSE= (SSxx * SSyy - SS²xy)/SSxx = 7317401.270

Explanation:

See the attached image file

Computers and Technology

Univariate linear regression Note: Solutions to this problem must follow the method described in class and the linear regression handout. There is some flexibility in how your solution is coded, but you may not use special functions that automatically perform linear regression for you. Load in the BodyBrain Weight.csv dataset. Perform linear regression using two different models: M1: brain_weight = w0 + w1 x body_weight M2: brain_weight = w0 + w1 x body_weight + w2 x body_weight2 a. For each model, follow the steps shown in class to solve for w.

Step-by-step answer

P Answered by Specialist

regression line is Y = 124.9281 + 0.9370 *x

SSE= (SSxx * SSyy - SS²xy)/SSxx = 7317401.270

Explanation:

See the attached image file

Business

The Grand Bakery produces 60 special sourdough rolls every day. Any rolls that are not sold each day are given to the employees. They have collected sales data from the past week: Day 1 50 rolls sold Day 2 50 rolls sold Day 3 48 rolls sold Day 4 60 rolls sold Day 5 53 rolls sold Day 6 60 rolls sold 1) What is the value of F6 if they use a 3-day weighted moving average with W1 = 0.6, W2 = 0.2, and W3 = 0.2? A. 51.4 B. 53.4 C. 58.6 D. None of the above 2) Using the data from the above, calculate the forecast for period 7 using a four-period moving

Step-by-step answer

P Answered by Specialist

B

D

Explanation:

The answers are 53.4 and 55.2 respectively. Please kindly go through the attached file for a run down and step by step solution to the question.

Business

The Grand Bakery produces 60 special sourdough rolls every day. Any rolls that are not sold each day are given to the employees. The bakery collected sales data from the past week: Day 1 50 rolls sold Day 2 50 rolls sold Day 3 48 rolls sold Day 4 60 rolls sold Day 5 53 rolls sold Day 6 60 rolls sold What is the value of F6 if they use a 3-day weighted moving average with W1 = 0.6, W2 = 0.2, and W3 = 0.2? A) 51.4 B) 53.4 C) 58.6 D) None of the above

Step-by-step answer

P Answered by PhD

The value of F6, if they use a 3- day weighted moving average is 53.4.

Explanation:

From the given data, the weighted moving average is

W1 = 0.6

W2 = 0.2

W3 = 0.2

To find the value of F6, the 3-day weighted moving average is used

The formula used to calculate is, where W is the weighted moving average and D is the days sold.

= W1 D1 + W2 D2 + W3 D3

= 0.6 (53) + 0.2(50) + 0.2(48)

= 53.4

Thus the value of F6 is 53.4.

Business

The Grand Bakery produces 60 special sourdough rolls every day. Any rolls that are not sold each day are given to the employees. They have collected sales data from the past week: Day 1 50 rolls sold Day 2 50 rolls sold Day 3 48 rolls sold Day 4 60 rolls sold Day 5 53 rolls sold Day 6 60 rolls sold 1) What is the value of F6 if they use a 3-day weighted moving average with W1 = 0.6, W2 = 0.2, and W3 = 0.2? A. 51.4 B. 53.4 C. 58.6 D. None of the above 2) Using the data from the above, calculate the forecast for period 7 using a four-period moving

Step-by-step answer

P Answered by Specialist

B

D

Explanation:

The answers are 53.4 and 55.2 respectively. Please kindly go through the attached file for a run down and step by step solution to the question.

Physics

Work and energy a 68 kg students runs up a flight of stairs 3.5 meters high in 11.4 seconds. the student takes 8.5 seconds to run up the same flight of stairs during a second trial. determine the following: a. the work done in each trial. you should identify these values as w1 and w2. b. the power exerted by the student in each trial. you should identify these values as p1 and p2.

Step-by-step answer

P Answered by PhD

1

a) W_1 = 2332 J , W_2= 2332 J

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

$W=mg\Delta h$

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the gain in height of the student

For the first student, $\Delta h = 3.5 m$ , so the work done is

W_1 = (68)(9.8)(3.5)=2332 J

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

W_2 = (68)(9.8)(3.5)=2332 J

2) P_1 = 204.6 W , P_2 = 274.4 W

The power exerted by each student is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time taken

For the first student, W_1 = 2332 J and t=11.4 s , so the power exerted is

$P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W$

For the second student, W_2 = 2332 J and t=8.5 s , so the power exerted is

$P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W$

Physics

Work and energy a 68 kg students runs up a flight of stairs 3.5 meters high in 11.4 seconds. the student takes 8.5 seconds to run up the same flight of stairs during a second trial. determine the following: a. the work done in each trial. you should identify these values as w1 and w2. b. the power exerted by the student in each trial. you should identify these values as p1 and p2.

Step-by-step answer

P Answered by PhD

1

a) W_1 = 2332 J , W_2= 2332 J

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

$W=mg\Delta h$

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the gain in height of the student

For the first student, $\Delta h = 3.5 m$ , so the work done is

W_1 = (68)(9.8)(3.5)=2332 J

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

W_2 = (68)(9.8)(3.5)=2332 J

2) P_1 = 204.6 W , P_2 = 274.4 W

The power exerted by each student is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time taken

For the first student, W_1 = 2332 J and t=11.4 s , so the power exerted is

$P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W$

For the second student, W_2 = 2332 J and t=8.5 s , so the power exerted is

$P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W$

4. Find the value of W2 if W1 = 250N.

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