N is an integer greater than 1.

A horse race has 1313 entries and one person owns 44 of those horses. Assuming that there are no ties, what is the probability

777dan777 [17]

Answer: 0.0014

Step-by-step explanation:

Given : Number of entries in horse race = 13

Number of horses owned = 4

We assume that there are no ties .

Then , the probability that those four horses finish first, second ,third ,and fourth is given by :-

$\dfrac{4!}{^{13}P_4}=\dfrac{4!}{\dfrac{13!}{(13-4)!}}\\\\=\dfrac{4!\times9!}{13!}=.0013986013986\approx0.0014$

Hence, the required probability = 0.0014

6 0

4 years ago

Which of the following is NOT classified as nonverbal communication a. Signs b. Eye Contact c. Body language d. Gestures

Dafna11 [192]

Answer:

american sign language (ASL)

4 0

3 years ago

Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The following ar

Softa [21]

Answer:

A. E ( U ) = 21.5454 , E ( F ) = 8.39333

B. M ( U ) = 17.0 , M ( F ) = 18.0

C. E ( U' ) = 17.0 , E ( F' ) = 7.95384

D. T ( U ) = 9.091% , T ( F ) = 6.667%

Step-by-step explanation:

Solution:-

- Two sample sets ( U ) and ( F ) that define the concentration ( EU/mg ) of endotoxin found in urban and farm homes as follows:

U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0

F: 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 22.0 9.6 2.0 2.0 0.5

- To determine the mean of a sample E ( U ) or E ( F ) the following formula from descriptive statistics is used:

$E ( X ) = Sum ( X_i ) / n$

Where,

Xi : Data iteration

n: Sample size

Therefore,

$E ( U ) = \frac{Sum (U_i )}{n_u} \\\\E ( U ) = \frac{6.0 + 5.0 + 11.0 + 33.0 + 4.0+ 5.0 +80.0+ 18.0+ 35.0+ 17.0+ 23.0 }{11} \\\\E ( U ) = 21.54545\\\\E ( F ) = \frac{Sum (F_i )}{n_f} \\\\E ( F ) = \frac{2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0+ 3.0+ 9.8+ 22.0+ 9.6+ 2.0+ 2.0+ 0.5 }{15} \\\\E ( F ) = 8.39333$

- To determine the sample median we need to arrange the data for both samples ( U ) and ( F ) in ascending order as follows:

U: 4.0 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0 80.0

F: 0.5 2.0 2.0 2.0 3.0 6.0 7.0 8.0 8.0 9.6 9.8 12.0 15.0 19.0 22.0

- Now find the mid value for both sets:

Median term ( U ) = ( n + 1 ) / 2

= ( 11 + 1 ) / 2 = 12/2 = 6th term

Median ( U ), 6th term = 17.0

Median term ( F ) = ( n + 1 ) / 2

= ( 15 + 1 ) / 2 = 16/2 = 8th term

Median ( F ), 8th term = 8.0

- We will now trim the smallest and largest observation from each set.

- In set ( U ) we see that smallest data corresponds to ( 4.0 ) while the largest data corresponds to ( 80.0 ). We will exclude these two terms and the trimmed set is defined as:

U': 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0

- In set ( F ) we see that the smallest data corresponds to ( 0.5 ) while the largest data corresponds to ( 22.0 ). We will exclude these two terms and the trimmed set is defined as:

F': 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 9.6 2.0 2.0

- We will again use the previous formula to calculate means of trimmed samples ( U' ) and ( F' ) as follows:

$E ( U' ) = \frac{5.0+ 5.0+ 6.0+ 11.0+ 17.0+ 18.0+ 23.0+ 33.0+ 35.0}{9} \\\\E ( U' ) = 17$

$E ( F' ) = \frac{2.0 +2.0+ 2.0 +3.0+ 6.0+ 7.0+ 8.0+ 8.0+ 9.6+ 9.8+ 12.0+ 15.0+ 19.0}{13} \\\\E ( F' ) = 7.95384$

- The trimming percentage is known as the amount of data removed from the original sample from top and bottom of sample size of 11 and 15, respectively.

- We removed the smallest and largest value from each set. Hence, a single value was removed from both top and bottom of each data set. We can express the trimming percentage for each set as follows:

$T ( U ) = \frac{1}{11} * 100 = 9.091\\\\T ( F ) = \frac{1}{15} * 100 = 6.667$ %