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Aleks04 [339]
3 years ago
11

Need help will give brianly

Mathematics
1 answer:
Katen [24]3 years ago
8 0

3. No it is not on the graph.

because: y=2x+5 insert (5,6)

              6=2x5 +5 is wrong

4. C

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3 tenths +4 tenths +__________tenths
algol [13]

Well four plus seven would equal seven tenths.

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3 years ago
Consider the original triangle and its enlargement.
Trava [24]

Answer:

I believe the scale factor is 2.

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An airplane travels at a constant speed of 300 miles per hour. How far, in miles, will the airplane travel in 20 minutes?​
Strike441 [17]

Answer:

100 miles in 20 minutes.

Step-by-step explanation:

300/60 = x/20

x = 300/60 * 20

x = 100

7 0
3 years ago
|5x|-2=18 helps love this problem please help
drek231 [11]

Answer:

x = 4

Step-by-step explanation:

The absolute value of 5x is still 5x.

5x - 2 = 18

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7 0
4 years ago
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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

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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