<span>1 x 42, 2 x 21, 3 x 14, or 6 x 7</span>
Answer:
25000000 is the correct answer
Using the hypergeometric distribution, it is found that there is a 0.0452 = 4.52% probability that this group of four students includes at least two of the top three geography students in the class.
The students are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.
<h3>What is the hypergeometric distribution formula?</h3>
The <em>formula </em>is:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7DC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 28 students in class, hence
.
- Four people will be chosen at random, hence
.
- The top three is composed by 3 people, hence
.
The probability that this group of four students includes at least two of the top three geography students in the class is:
![P(X \geq 2) = P(X = 2) + P(X = 3)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29)
In which:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7DC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![P(X = 2) = h(2,28,4,3) = \frac{C_{3,2}C_{25,2}}{C_{28,4}} = 0.0440](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20h%282%2C28%2C4%2C3%29%20%3D%20%5Cfrac%7BC_%7B3%2C2%7DC_%7B25%2C2%7D%7D%7BC_%7B28%2C4%7D%7D%20%3D%200.0440)
![P(X = 3) = h(3,28,4,3) = \frac{C_{3,3}C_{25,1}}{C_{28,4}} = 0.0012](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20h%283%2C28%2C4%2C3%29%20%3D%20%5Cfrac%7BC_%7B3%2C3%7DC_%7B25%2C1%7D%7D%7BC_%7B28%2C4%7D%7D%20%3D%200.0012)
Then:
![P(X \geq 2) = P(X = 2) + P(X = 3) = 0.0440 + 0.0012 = 0.0452](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%3D%200.0440%20%2B%200.0012%20%3D%200.0452)
0.0452 = 4.52% probability that this group of four students includes at least two of the top three geography students in the class.
You can learn more about the hypergeometric distribution at brainly.com/question/4818951
Answer:
![Area = 538.5 m^2](https://tex.z-dn.net/?f=%20Area%20%3D%20538.5%20m%5E2%20)
Step-by-step Explanation:
Given:
∆XVW
m < X = 50°
m < W = 63°
XV = w = 37 m
Required:
Area of ∆XVW
Solution:
<em>Find side length XW using Law of Sines</em>
![\frac{v}{sin(V)} = \frac{w}{sin(W)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%7D%7Bsin%28V%29%7D%20%3D%20%5Cfrac%7Bw%7D%7Bsin%28W%29%7D%20)
W = 63°
w = XV = 37 m
V = 180 - (50+63) = 67°
v = XW = ?
![\frac{v}{sin(67)} = \frac{37}{sin(63)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%7D%7Bsin%2867%29%7D%20%3D%20%5Cfrac%7B37%7D%7Bsin%2863%29%7D%20)
Cross multiply
![v*sin(63) = 37*sin(67)](https://tex.z-dn.net/?f=%20v%2Asin%2863%29%20%3D%2037%2Asin%2867%29%20)
Divide both sides by sin(63) to make v the subject of formula
![\frac{v*sin(63)}{sin(63)} = \frac{37*sin(67)}{sin(63)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%2Asin%2863%29%7D%7Bsin%2863%29%7D%20%3D%20%5Cfrac%7B37%2Asin%2867%29%7D%7Bsin%2863%29%7D%20)
![v = \frac{37*sin(67)}{sin(63)}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B37%2Asin%2867%29%7D%7Bsin%2863%29%7D%20)
(approximated to nearest whole number)
![XW = v = 38 m](https://tex.z-dn.net/?f=%20XW%20%3D%20v%20%3D%2038%20m%20)
<em>Find the area of ∆XVW</em>
![area = \frac{1}{2}*v*w*sin(X)](https://tex.z-dn.net/?f=%20area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2Av%2Aw%2Asin%28X%29%20)
![= \frac{1}{2}*38*37*sin(50)](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A38%2A37%2Asin%2850%29%20)
![= \frac{38*37*sin(50)}{2}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B38%2A37%2Asin%2850%29%7D%7B2%7D%20)
(to nearest tenth).
The answer to the first question is -6/13