4,133,000 2,369,000 1,295,000 928,000 679,000

Round your answers to the nearest integer.

a. The computations will be easier to work if you view this problem in terms

of thousands of passengers. Represent each number in terms of thousands

of passengers.

b. What is the mean number of passengers for these five cruise lines? (Give

the full number.)

c. What is the range? (Give the full number.)

d. What is the standard deviation? (Give the full number.)

Answer:
### Final answer:

### Explanation:

### Learn more about Statistics here:

The numbers in thousands are: 4133, 2369, 1295, 928, 679. The mean is 1881 thousand passengers, the range is 3454 thousand passengers, and the standard deviation is approximately 1218 thousand passengers.

**Part A:** To represent each number in terms of thousands of passengers, we simplify them as follows: 4,133,000 is 4,133 thousands of passengers, 2,369,000 is 2,369 thousands, 1,295,000 is 1,295 thousands, 928,000 is 928 thousands, and 679,000 is 679 thousands.

**Part B:** To calculate the mean number of passengers, you would add up all the passenger numbers and then divide by the number of values (5 in this case). This adds up to 9,404,000 passengers or, in terms of thousands, 9,404. Divided by 5, this gives a mean of 1,880,800 passengers, or 1,881 thousands of passengers.

**Part C:** The range is calculated by subtracting the smallest number from the largest. In this case, that would be 4,133,000 - 679,000 = 3,454,000, which is also 3,454 thousands of passengers.

**Part D:** Calculating the standard deviation involves multiple steps. First, for each value, subtract the mean and square this result. Then, calculate the mean of these squared differences. Finally, take the square root of this mean. Doing so, the standard deviation is approximate 1,217,982 passengers, or 1,218 thousands of passengers.

#SPJ12

To make the cone, melted chocolate is pumped onto a huge cold plate at a rate of 2 ft3/sec. Due to the low temperature, the chocolate forms the shape of a cone as it solidifies quickly. If the height is always equal to the diameter as the cone is formed, how fast is the height of the cone changing when it is 5 ft high?

Bryan is a hotel manager.his salary is 7500. Every year his salary increases by 150 what will be his salary in 5 years

Below is a list of the scores of the Altanta Falcons games scores in the year 2021. 3 17 10 6 25 17 34 30 13 27 3 0 21 17 13 20 15 20 29 27What is the mean score?

What is (1/3) x (-2/3 )?

For a field trip 9 students rode in cars andthe rest filled eight buses. How many students were in each bus if 265 students were on the trip?

Bryan is a hotel manager.his salary is 7500. Every year his salary increases by 150 what will be his salary in 5 years

Below is a list of the scores of the Altanta Falcons games scores in the year 2021. 3 17 10 6 25 17 34 30 13 27 3 0 21 17 13 20 15 20 29 27What is the mean score?

What is (1/3) x (-2/3 )?

For a field trip 9 students rode in cars andthe rest filled eight buses. How many students were in each bus if 265 students were on the trip?

**Step-by-step explanation:**

f is a parabola that opens upward.

**Answer:**

The probability that he or she is high-risk is **0.50**

**Step-by-step explanation:**

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to calculate P( High risk | at-fault accident) = ?

**Using Bayes' conditional probability theorem**

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) / { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) + P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50.

The **probability** that a driver is high-risk given that they had an at-fault accident can be found using Bayes' theorem. Given the probabilities provided in the question, the probability is approximately 0.3333 or 33.33%.

To find the probability that a driver is high-risk given that they had an at-fault accident, we can use Bayes' theorem. Let's define the events:

- A: Driver is high-risk
- B: Driver has an at-fault accident

We are given the following probabilities:

- P(A) = 0.20 (probability of a driver being high-risk)
- P(B|A) = 0.20 (probability of an at-fault accident given that they are high-risk)
- P(~A) = 0.80 (probability of a driver not being high-risk)
- P(B|~A) = 0.10 (probability of an at-fault accident given that they are not high-risk)

Using Bayes' theorem, the probability of a driver being high-risk given that they had an at-fault accident is:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(~A) * P(B|~A))

Substituting the given probabilities:

P(A|B) = (0.20 * 0.20) / (0.20 * 0.20 + 0.80 * 0.10) = 0.04 / (0.04 + 0.08) = 0.04 / 0.12 = 0.3333.

Therefore, the **probability** that a driver is high-risk given that they had an at-fault accident in the next year is approximately 0.3333 or 33.33%.

#SPJ3

**Answer:**

2

**Step-by-step explanation:**

it's division 10/5 is 2 which means in one second it travels 2 floors

answer: 2

5/5 = 1

10/5 = 2

2 floors/second

5/5 = 1

10/5 = 2

2 floors/second

**Answer:**

Nancy put 3 cups of vegetables in each container.

**Step-by-step explanation:**

1. Add the amount of vegetables together

2. Divided the number of vegetables by 5, since they are put into 5 separate containers.

____________________________________________________________2+1+2+1+3+6 = 15

15 divided by 5 or 5 x ? = 15

**Answer:**

2 1/2 cups of vegetables

**Step-by-step explanation:**

This is an averages question, so add all numbers and divide by amount of numbers

2+1+2+1+3+6=15

15/6=2 1/2

Part B: Write the equation obtained in Part A using function notation.

Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals.

First find the slope using the two points (2,225) and (5,480)

(480-225)/(5-2) = 85

Then find b.

225 = 85*2 + b

225 = 170 + b

55 = b

y = 85x + 55 OR f(x) = 85x + 55 This is the answer to PART B.

Standard form PART A is 85x - y = 55

PART C Plot the y intercept, 55. From the y intercept count the slope 85 up right 1 and plot a point, draw a line through the points

The x axis is the number of days so it would be in intervals of 1.

The y axis is money, I would do intervals of $5

(480-225)/(5-2) = 85

Then find b.

225 = 85*2 + b

225 = 170 + b

55 = b

y = 85x + 55 OR f(x) = 85x + 55 This is the answer to PART B.

Standard form PART A is 85x - y = 55

PART C Plot the y intercept, 55. From the y intercept count the slope 85 up right 1 and plot a point, draw a line through the points

The x axis is the number of days so it would be in intervals of 1.

The y axis is money, I would do intervals of $5

24 were Bb

21 were bb

Assume that this is a random sample.

A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes. The procedure obtained P = 0.0001.

1. "The frequency distribution of genotypes has a binomial distribution in the population" is the________ hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the _________ hypothesis.

2. The degrees of freedom for the test statistic are __________.

Say whether the each of the following statements is true or false solely on the basis of these results:

3. The difference between the observed and expected frequencies is statistically significant.________

4. The test statistic exceeds the critical value corresponding to α = 0.05. ___________

5. The test statistic exceeds the critical value corresponding to α = 0.01. ____________

**Answer:**

1) i ) N**ull hypothesis,** ii) **Alternate hypothesis**

2) Degree of Freedom =** 14.9011**

3) **True**

4) **True**

5) **True**

**Step-by-step explanation:**

1) "The frequency distribution of genotypes has a binomial distribution in the population" is the __ NULL__ hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the

2) **Degrees of freedom**: see attached calculation for

3) True, the difference between the observed and expected frequencies is statistically significant since **p-value** < ** ∝****alpha(0.05)**

4) True

**when**, 14.9011 > 3.8415 where ∝ = 0.05

**when**, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.05

5) True

**when**, 14.9011 > 6.635 where ∝ = 0.01

**when**, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.01

The** null hypothesis** states that the genotype frequency distribution follows a binomial distribution, with the alternative hypothesis stating the contrary. The **degrees of freedom** for the **test statistic** are two. The P value of 0.0001 is statistically significant, thus the test statistic exceeds the critical values corresponding to both α = 0.05 and α = 0.01.

1. **The frequency distribution of genotypes has a binomial distribution in the population** is the **null hypothesis**, whereas *The frequency distribution of genotypes does not have a binomial distribution* is the** alternative hypothesis**.

2. For this chi-square test, since there are 3 possible outcomes (BB, Bb, bb) the degrees of freedom are 3-1 =** 2**.

3. The difference between the observed and expected frequencies is statistically significant, that is **true**. A P value of 0.0001 is highly significant, clearly less than 0.05, and denotes a significant difference.

4. The test statistic exceeds the critical value corresponding to α = 0.05. This is **true**, as the statistically significant low P value indicates the test statistic is in the critical region.

5. The test statistic exceeds the critical value corresponding to α = 0.01. This is also **true**.

#SPJ11