Questions
A heavy semi-truck collides head-on with a small, light car. Before the collision the vehicles were...

A heavy semi-truck collides head-on with a small, light car. Before the collision the vehicles were moving with the same speed but in opposite direction.

A. Just before the collision, was the magnitude of the momentum of the car greater, equal or less than that of the truck?

B. Just before the collision, was the kinetic energy of the car greater, equal or less than that of the truck?

C. During the collision, is the force that the car exerts on the truck greater, equal or less than the force that the truck exerts on the car?

D. During the collision, is the magnitude of the acceleration of the car greater, equal or less than the magnitude of the acceleration of the truck?

E. If the two vehicles stick together immediately after the collision, which way will they be moving, if at all?

In: Physics

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly...

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim?

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.

Golfer 1 2 3 4 5 6 7 8
Score (old design) 94 73 90 82 85 76 81 79
Score (new design) 88 80 88 75 86 72 80 75

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Find the p-value for the hypothesis test. Round your answer to four decimal places.

Step 5 of 5: Draw a conclusion for the hypothesis test.

In: Statistics and Probability

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly...

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim?

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.

Golfer 1 2 3 4 5 6 7 8
Score (old design) 77 88 76 88 84 80 89 81
Score (new design) 71 92 73 87 91 73 83 78

Step 1 of 5:

State the null and alternative hypotheses for the test.

Step 2 of 5:

Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:

Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5:

Find the p-value for the hypothesis test. Round your answer to four decimal places.

Step 5 of 5:

Draw a conclusion for the hypothesis test.

In: Statistics and Probability

Use Einstein's velocity addition rule and the principle of momentum conservation to explain why you can't...

Use Einstein's velocity addition rule and the principle of momentum conservation to explain why you can't use the formula p=mv to find the momentum of an object that is moving relativistically, and show that the correct formula for relativistic momentum is actually p = γmv. Use the following bullet points as a guide for what points to cover in your essay. First imagine you have two objects, A and B, that are moving in the same direction relative to you but with different speeds. Object B is ahead of object A, but the separation between them is narrowing because object A is faster. Now suppose object A catches up to object B and they collide. They keep going in the same direction after the collision but with new speeds. What is the total non-relativistic momentum before and after the collision? Now suppose the collision is taking place aboard a moving train that you are observing from the ground. Let u be the speed of the train relative to you. Use Einstein's velocity addition rule to add the train's motion to the motions of objects A and B to find their speeds relative to you. What is the total non-relativistic momentum before and after the collision, now with the motion of the train taken into account? If you set them equal to each other, does the speed of the train u just cancel out? You should have found that when you use the non-relativistic formula for momentum, the train motion did not cancel out. Explain why this implies that non-relativistic momentum is not conserved on the moving train. Now show that u does cancel out when you use the relativistic formula for total momentum before and after the collision between objects A and B on the moving train. Explain why this implies that relativistic momentum is conserved.

In: Physics

On June 30, 2017, Sharper Corporation’s common stock is priced at $27.00 per share before any...

On June 30, 2017, Sharper Corporation’s common stock is priced at $27.00 per share before any stock dividend or split, and the stockholders’ equity section of its balance sheet appears as follows. Common stock—$6 par value, 90,000 shares authorized, 36,000 shares issued and outstanding $ 216,000 Paid-in capital in excess of par value, common stock 100,000 Retained earnings 316,000 Total stockholders’ equity $ 632,000

1. Assume that the company declares and immediately distributes a 100% stock dividend. This event is recorded by capitalizing retained earnings equal to the stock’s par value. Answer these questions about stockholders’ equity as it exists after issuing the new shares. a.,b.& c. Complete the below table to calculate the retained earnings balance, total stockholders’ equity and number of outstanding shares.

Stock Dividend Before Stock Dividend Impact of Stock Dividend After Stock Dividend
Common stock
Paid in capital in excess of par value
Total contributed capital
Retained earnings
Total stockholders' equity
Number of common shares outstanding

2. Assume that the company implements a 2-for-1 stock split instead of the stock dividend in part 1.

Answer these questions about stockholders’ equity as it exists after issuing the new shares. a.,b.& c. Complete the below table to calculate the retained earnings balance, total stockholders’ equity and number of outstanding shares.

Stock Split Before Stock Split Impact of Stock Split After Stock Split
Common stock
Paid in capital in excess of par value
Total contributed capital 0 0
Retained earnings
Total stockholders' equity $0 $0
Number of common shares outstanding

In: Accounting

Dependent Sample T-Test Researchers were interested in studying whether seniors who participated in Silver Sneakers had...

Dependent Sample T-Test

Researchers were interested in studying whether seniors who participated in Silver Sneakers had a significant weight loss over a 10 week period. Given the above information, answer the questions below (assume a level of significance of .05).

  1. State the null and alternate hypotheses.
  2. What is the independent variable?
  3. What is the dependent variable?
  4. Complete the table below:

Paired Samples Statistics

Mean

N

Std. Deviation

Std. Error Mean

Pair 1

preweight Weight before the diet (kg)

72.53

78

8.723

.988

weight10weeks Weight after 10 weeks on the diet

68.681

78

8.9245

1.0105

Beginning Weight in kg

Weight after 10 weeks in kg

Sample Size

Sample Mean

Sample Standard Deviation

Estimated Standard Error

Using the information below, answer the following questions:

  1. What is the mean difference?
  2. What was the t value and degrees of freedom?
  3. What was the p value?
  4. Draw a conclusion based on the hypotheses. Phrase it in terms of the original research questions, not just whether you accept or reject the null.

Paired Samples Test

Paired Differences

Mean

Std. Deviation

Std. Error Mean

95% Confidence Interval of the Difference

Lower

Pair 1

Weight before the diet (kg) - Weight after 10 weeks on the diet

3.8449

2.5515

.2889

3.2696

Paired Samples Test

Paired Differences

t

df

Sig. (2-tailed)

95% Confidence Interval of the Difference

Upper

Pair 1

Weight before the diet (kg) - weight10weeks Weight after 10 weeks on the diet

4.4201

13.309

77

.000

In: Statistics and Probability

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly...

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim?

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.

Golfer 1 2 3 4 5 6 7 8
Score (old design) 77 76 95 77 80 81 85 79
Score (new design) 73 83 94 73 81 76 79 73

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5:Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5:Determine the decision rule for rejecting the null hypothesis H 0  H0 . Round the numerical portion of your answer to three decimal places.

Step 5 of 5: accept or reject null?

In: Statistics and Probability

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly...

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim?

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.1α=0.1 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.

Golfer 1 2 3 4 5 6 7 8
Score (old design) 89 83 85 95 89 83 86 89
Score (new design) 86 90 78 89 95 82 85 85

Step 1 of 5:State the null and alternative hypotheses for the test.

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Find the p-value for the hypothesis test. Round your answer to four decimal places.

Step 5 of 5: Draw a conclusion for the hypothesis test

In: Statistics and Probability

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly...

A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim?

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.

Golfer 1 2 3 4 5 6 7 8
Score (old design) 93 85 78 75 84 75 81 77
Score (new design) 90 87 75 71 91 71 79 75

Step 1: State the null and alternative hypotheses for the test.

Step 2: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5: Make the decision for the hypothesis test.

In: Statistics and Probability

A $10 000 bond with 5% interest payable quarterly, redeemable at par on November 15, 2030,...

A $10 000 bond with 5% interest payable quarterly, redeemable at par on November 15, 2030, was bought on July 2, 2014, to yield 9% compounded quarterly. If the bond sells at 92.75 on September 10, 2020, what would the gain or loss on the sale be?

   Face value = 10 000.00; b = 1.25%

                   Principal = 10 000.00; i = 2.25%

                   Interest dates are November 15, February 15, May 15, and August 15.

                   The interest date preceding the date of sale is August 15, 2020.

                   The time period August 15, 2020, to November 15, 2030, is 10.25 years: n = 41.

                   b < i → discount

                   The interest payment interval August 15, 2020, to November 15, 2020, is 92 days

                   The interest period August 15, 2020, to September 10, 2020, is 26 days.

In: Finance