Billiard ball A of mass mA = 0.122 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.145 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s, and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.

1) Solve these equations for the angle, θ′B, of ball B after the collision. Do not assume the collision is elastic.

2)Solve these equations for the speed, v′B, of ball B after the collision. Do not assume the collision is elastic.

Consider three gas connected gas containers as shown in the figure below.

Consider three gas connected gas containers as shown in the figure below.

CO2 (g) P = 2.13, V = 1.50L

H2(g) P = 0.861 atm, V = 1.00 L

Ar (g) p = 1.15atm, V = 2.00L

   T = 298K

Assume, that you open the two stopcock so that the gases can flow freely.

Calculate the change of the chemical potential of the entire system due to mixing the gases.

Using the change of the chemical potential of gases show that the mixing of three gases is spontaneous

Determine the value of D_mixH .

_mix = ______________________

_mixH = _____________________

1. Consider the following expression:

Everyone who is nice to all dogs is loved by somebody.

∀x (∀y dog(y) ⇒ Kind(x,y)) ⇒ (∃y Loves(y,x))

Demonstrate how to convert it to CNF. You should get:
( dog(F(x)) v (Loves(G(x),x)) ^ (~nice(x,F(x)) v Loves(G(x),x))

Step 1: Remove the biconditionals and implications.

Step 2: Move the negations inward.

Step 3: Standardize variables

Step 4: Skolemize

Step 5: Drop universal quantifiers

Step 6: Distribute v over ^

Expand 1.00 mol of a monatomic gas, initially at 3.60 kPa and 313 K, from initial volume V_i = 0.723 m³ to final volume V_f = 2.70 m³. At any instant during the expansion, the pressure p and volume V of the gas are related by p = 3.60 exp[(V_i - V)/a], with p in kilopascals, V_i and V are in cubic meters, and a = 2.10 m³. What are the final (a) pressure and (b)temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is the change in entropy of the gas for the expansion? (Hint: Use two simple reversible processes to find the entropy change.)

Requirements: Code in C++. With given information, write the solution to this problem so that it is understandable to someone with basic knowledge of C++ (ex: only keep basic libraries, keep coding shortcuts to a minimum). Also leave comments in the code (plz), the logic behind solving this problem if possible, and explanation of what the keys to solving this problem is and how to run test cases to ensure correctness of code.

Problem:

For this problem you will compute various running sums of values for positive integers.

Input

The first line of input contains a single integer P, (1 <=P <=10000), which is the number of data sets that follow. Each data set should be processed identically and independently. Each data set consists of a single line of input. It contains the data set number, K, followed by an integer N, (1 <= N <= 10000).

Output

For each data set there is one line of output. The single output line consists of the data set number, K, followed by a single space followed by three space separated integers S1, S2 and S3 such that: S1 = The sum of the first N positive integers. S2 = The sum of the first N odd integers. S3 = The sum of the first N even integers.

Sample Input

3

1 1

2 10

3 1001

Sample Output

1 1 1 2

2 55 100 110

3 501501 1002001 1003002  

A Solution that needs to be rewritten so it can be understood more easily

/*
 * Sum Kind Of Problem
 * Compute running sums of odds, evens and all integers
 * Does it 2 ways... using formulas and loops.
 * Define EASY_WAY for formulas
 * Define HARD_WAY for loops
 * Define both of them to show both results
 */
#include <stdio.h>
#include <stdlib.h>

#define EASY_WAY
#undef LONG_WAY

int main()
{
        int n, i, v, k;
#ifdef LONG_WAY
        int oddsum, evensum, sum, j;
#endif

        scanf("%d", &n);
        for(i = 1; i <= n; i++){
                scanf("%d %d", &(k), &(v));
#ifdef EASY_WAY
                printf("%d %d %d %d\n", i, v*(v+1)/2, v*v, v*(v+1));
#endif
#ifdef LONG_WAY
                oddsum = evensum = sum = 0;
                for(j = 1; j <= v; j++){
                        evensum += j*2;
                        oddsum += j*2 - 1;
                        sum += j;
                }
                printf("%d %d %d %d\n", i, sum, oddsum, evensum);
#endif
        }
        return(0);
}

The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $23,300. You take a random sample of 136 college students in the state of Vermont and find the mean debt is $24,500 with a standard deviation of $2,700. You want to construct a 99% confidence interval for the mean debt for all Vermont college students.

(a) What is the point estimate for the mean debt of all Vermont college students?

(b) Construct the 99% confidence interval for the mean debt of all Vermont college students. Round your answers to the nearest whole dollar.

(c) Are you 99% confident that the mean debt of all Vermont college students is greater than the quoted national average of $23,300 and why?

a. Yes, because $23,300 is above the lower limit of the confidence interval for Vermont students.

b. No, because $23,300 is above the lower limit of the confidence interval for Vermont students.

c. No, because $23,300 is below the lower limit of the confidence interval for Vermont students.

d. Yes, because $23,300 is below the lower limit of the confidence interval for Vermont students.

d) We are never told whether or not the parent population is normally distributed. Why could we use the above method to find the confidence interval?

a. Because the margin of error is positive.

b. Because the margin of error is less than 30.

c. Because the sample size is greater than 30.

d. Because the sample size is less than 100.

A researcher at the college used a random sample of 30 students to investigate sleep patterns of two groups. The output below was constructed from this study for comparing average sleep hours of students who have early classes versus students who do not have early classes.

The hours of sleep for both group of students are stored in EarlyClass.xlsx. Download and open this file into Excel. Use Excel to undertake a suitable test to address the research question below.

Is there a difference between the average hours of sleep for students with early classes versus students with no early classes?

Answer the following questions by choosing the most correct option or typing the answer:

1. (1 mark) The most appropriate test for these data is: AnswerA two-sample t-testA paired t-test

2. (1 mark) The normality assumption seems reasonable because: Answerboth sleep hours for students with early classes and no early classes may be drawn from normal distributions.the differences between sleep hours of students with no early classes and with early classes may be drawn from a normal distribution.

3. (1 mark) The assumption of equal variance seems reasonable because: Answercomparative boxplots suggest the variation in the two populations could be the same.comparative boxplots suggest the variation in the two populations is different.

For the remaining questions you may assume that any relevant assumptions have been met.

4. The absolute value of the test statistic is equal to (type your answer with 3 dp)  Answer

5. (1 mark) The degrees of freedom is equal to (type your answer as an integer)  Answer

6. (1 mark) The p-value is larger than 0.05 Answertruefalse

7. The test shows that the average hours of sleep was Answersignificantly greater for students with no early classes than for students with early classes.significantly greater for students with early classes than for students with no early classes.not significantly different for students with no early classes and for students with early classes.

please include working out if possible

A researcher wants to determine whether high school students who attend an SAT preparation course score significantly different on the SAT than students who do not attend the preparation course. For those who do not attend the course, the population mean is 1050 (μ = 1050). The 16 students who attend the preparation course average 1150 on the SAT, with a sample standard deviation of 300. On the basis of these data, can the researcher conclude that the preparation course has a significant difference on SAT scores? Set alpha equal to .05.

Q1: The appropriate statistical procedure for this example would be a

A. z-test

B. t-test

Q2: The most appropriate null hypothesis (in words) would be

A. There is no statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.

B. There is a statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.

C. The students who took the SAT prep course did not score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.

D. The students who took the SAT prep course did score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.

Q3: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

B. μSATprep = 1150

C. μSATprep  1050

D. μSATprep  1050

Q4: Based on your evaluation of the null in and your conclusion, as a researcher you would be more concerned with a

A. Type I statistical error

B. Type II statistical error

Calculate the 99% confidence interval. Steps:

Q5: The mean you will use for this calculation is

A. 1050

B. 1150

Q6: What is the new critical value you will use for this calculation?

Q7: As you know, two values will be required to complete the following equation:

__________    __________

Q8: Which of the following is a more accurate interpretation of the confidence interval you just computed?

A. We are 99% confident that the scores fall in the interval _____ to _____.

B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval _____ to _____.

C. We are 99% confident that the example above has correct values.

D. We are 99% confident that the difference in SAT scores between the students who took the prep course and the students who did not falls in the interval _____ to _____.