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tangare [24]
3 years ago
8

Given the polynomial 6x3 + 4x2 − 6x − 4, what is the value of the coefficient 'k' in the factored form? 6x3 + 4x2 − 6x − 4 = 2(x

+ k)(x − k)(3x + 2) k= ____________ Numerical Answers Expected! Answer for Blank 1:
Mathematics
1 answer:
anastassius [24]3 years ago
5 0

6x^3+4x^2-6x-4 =\\ 2(3x^3+2x^2-3x-2) =\\ 2(x^2(3x+2)-1(3x+2))=\\ 2(x^2-1)(3x+2)=2(x+1)(x-1)(3x+2).

So k=1.

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Which number is not equivalent to others? 0.5 1/5 5/25 20%
CaHeK987 [17]

Answer: .5

Step-by-step explanation:

1/5 = 20%

5/25=20%

.5= 50%

Hope this helps!

6 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}

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3 years ago
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emmasim [6.3K]
Use desmo will help you to find the answer
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Citrus2011 [14]

Answer:2 meters per second.

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C

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