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oksian1 [2.3K]
3 years ago
12

HELP HELP HELPPPP ILL GIVE BRAINLIEST HELPPPPPPPPP 100 POINTSSS

Mathematics
2 answers:
ahrayia [7]3 years ago
7 0

Answer:

C. 0.48

Step-by-step explanation:

Probability = number of required outcome

_______________________

number of possible outcome

= total volleyball game events

_______________________

total sophomore + junior

= 66/137

= 0.48

omeli [17]3 years ago
4 0

Answer: D) 0.31

Step-by-step explanation:

Let A denote the event that a person is a sophomore.

Let B denote the event that a person has attended volleyball game.

A∩B denote the event that a person is a sophomore and attend volleyball game.

Let P denote the probability of an event.

We are asked to find:

P(A∩B)

From the table provided to us we see that:

A∩B=42

Hence,

P(A∩B)=42/137=0.3065 which is approximately equal to 0.31. Therefore ur answer will be 0.31.

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Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first,
Kryger [21]

Answer:

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value x \in (0,1) we want this:

max(U_1, ....,U_n) \leq x

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u_i \leq x for each possible i

We assume that the random variable u_i are independent and P)U_i \leq x) =x from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n

And then cumulative distribution would be expressed like this:

0, x \leq 0

x^n, x \in (0,1)

1, x \geq 1

For each value x\in (0,1) we can find the dendity function like this:

f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}

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f_X (x) = n x^{n-1} , x \in (0,1)  and 0 for other case

And now we can find the expected value for the random variable X like this:

E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

6 0
3 years ago
Help if you know thanks
Yanka [14]

Answer:

x = -1/2  x=-1

Step-by-step explanation:

2x( x+1.5) = -1

Distribute

2x^2 + 3x = -1

Add 1 to each side

2x^2 +3x+1 = 0

Factor

(2x+1) (x+1) =0

Using the zero product property

2x+1 = 0   x+1=0

2x = -1    x=-1

x = -1/2  x=-1

7 0
3 years ago
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What is the surface area of an orange with a 3inch radius
svp [43]

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3 0
2 years ago
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Answer:

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