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3 years ago

What ordered pair is missing from the input-output table?

Mathematics

2 answers:

AlexFokin [52]3 years ago

7 0

Answer:

(5,15)

Step-by-step explanation: to me it is just obvious

konstantin123 [22]3 years ago

5 0

Answer:

(5, 15)

Step-by-step explanation:

(X, Y) so if the x is 5 then 5 replaces x and if the y is 15 the 15 replaces y. So it would be (5,15)

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Licemer1 [7]

$\Bbb E[Y^2] = \Bbb E[(4X - 2)^2]$

so by definition of expectation,

$\Bbb E[Y^2] = \displaystyle \int_{-\infty}^\infty (4x-2)^2 f_X(x) \, dx = \int_0^\infty 3(4x-2)^2 e^{-3x} \, dx$

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$\displaystyle \int_a^b u\,dv = uv\bigg|_a^b - \int_a^b v\,du$

First, let

$u = (4x-2)^2 \implies du = 8(4x-2)\,dx \\\\ dv = 3e^{-3x} \, dx \implies v = -e^{-3x}$

so that

$\displaystyle \Bbb E[Y^2] = -(4x-2)^2 e^{-3x} \bigg|_{x=0}^{x\to\infty} + 8 \int_0^\infty (4x-2) e^{-3x} \, dx \\\\ ~~~~ = 4 + 8 \int_0^\infty (4x-2) e^{-3x} \, dx$

Next,

$u = 4x-2 \implies du = 4\,dx \\\\ dv = e^{-3x} \, dx \implies v = -\dfrac13 e^{-3x}$

so that

$\displaystyle \Bbb E[Y^2] = 4 + 8 \left(-\frac13 (4x-2) e^{-3x} \bigg|_{x=0}^{x\to\infty} + \frac43 \int_0^\infty e^{-3x} \, dx\right) \\\\ ~~~~ = 4 + 8 \left(-\frac23 - \frac49 e^{-3x}\bigg|_{x=0}^{x\to\infty}\right) \\\\ ~~~~ = 4 - \frac{16}3 + \frac{32}9 = \boxed{\frac{20}9}$

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1 year ago

Factor completely 2x3 + x2 - 18x - 9

melisa1 [442]

Factor the following:
2 x^3 + x^2 - 18 x - 9

Factor terms by grouping. 2 x^3 + x^2 - 18 x - 9 = (2 x^3 + x^2) + (-18 x - 9) = x^2 (2 x + 1) - 9 (2 x + 1):
x^2 (2 x + 1) - 9 (2 x + 1)

Factor 2 x + 1 from x^2 (2 x + 1) - 9 (2 x + 1):
(2 x + 1) (x^2 - 9)

x^2 - 9 = x^2 - 3^2:
(2 x + 1) (x^2 - 3^2)

Factor the difference of two squares. x^2 - 3^2 = (x - 3) (x + 3):

Answer: (x - 3) (x + 3) (2 x + 1) thus the Answer is C.

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