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diamong [38]
3 years ago
11

Can some one answer this

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
7 0

Answer:

Step-by-step explanation

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How put a number as a percentage
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Answer:

You need more information. But typically multiplly by 100 so .1 would be 10 %

Step-by-step explanation:


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Jack leased a car for $159/month for 36 months. The amount due when he signed the lease was $2399. How much did Jack pay during
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Quick question before I answer: Did he have to pay $2,399 when he signed the lease, or was that just how much he had to pay in total over the course of 36 months?
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A supervisor records the repair cost for 17 randomly selected VCRs. A sample mean of $68.83 and standard deviation of $12.17 are
amid [387]

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27

Step-by-step explanation:

im smart

5 0
3 years ago
2x+4y=24<br> -12x+8y=-16
EleoNora [17]
Dividing the first equation by 2
X + 2y = 12
Dividing the equation 2 by 4
-3x + 2y = -4
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4x = 16
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Also 4 + 2y = 12
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Y = 4
Please mark as brainliest
3 0
3 years ago
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

And we can express this like that:

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}

So then we have the pdf defined, and given by:

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
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