9514 1404 393
Answer:
(d) ∠AIS
Step-by-step explanation:
The tangent SF intersects the circle at one point, point I. Any angle it forms with a chord must be a chord that terminates at point I. There are two of them: AI and CI.
The angles these chords make with tangent SF are ...
Of these, only the first (∠AIS) is listed among the answer choices.
Using the hypergeometric distribution, there is a 0.4894 = 48.94% probability of selecting none of the correct six integers in a lottery.
<h3>What is the hypergeometric distribution formula?</h3>
The formula 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.
For this problem, we want to take 6 numbers from a set of 56, hence the values of the parameters are:
N = 56, k = 6, n = 6.
The probability of selecting none of the correct six integers in a lottery is of P(X = 0), hence:
![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 = 0) = h(0,56,6,6) = \frac{C_{6,0}C_{50,6}}{C_{56,6}} = 0.4894](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20h%280%2C56%2C6%2C6%29%20%3D%20%5Cfrac%7BC_%7B6%2C0%7DC_%7B50%2C6%7D%7D%7BC_%7B56%2C6%7D%7D%20%3D%200.4894)
0.4894 = 48.94% probability of selecting none of the correct six integers in a lottery.
More can be learned about the hypergeometric distribution at brainly.com/question/24826394
#SPJ4
Answer:
a) 0.857
b) 0.571
c) 1
Step-by-step explanation:
18 juniors
10 seniors
6 female seniors
10-6 = 4 male seniors
12 junior males
18-12 = 6 junior female
6+6 = 12 female
4+12 = 16 male
A total of 28 students
The possibility of each union of events is obtained by summing the probabilities of the separated events and substracting the intersection. I will abbreviate female by F, junior by J, male by M, senior by S; So we have
P(J U F) = P(J) + P(F) - P(JF) = 18/28+12/28-6/28 = 24/28 = 0.857
P(S U F) = P(S) + P(F) - P(SF) = 10/28 + 12/28 - 6/28 = 16/28 = 0.571
P(J U S) = P(J) + P(S) - P(JS) = 18/28 + 10/28 - 0 = 1
Note that a student cant be Junior and Senior at the same time, so the probability of the combined event is 0. The probability of the union is 1 because every student is either Junior or Senior.
(Extra; but please give me brainliest