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Degger [83]
3 years ago
13

Use distributive property to rewrite 6(11+x)=

Mathematics
2 answers:
mamaluj [8]3 years ago
8 0
Hey there!
The distributive property is defined as:
a(b+c) = ab+ac
Using this property, we can solve this problem.
Because it's a times c, we first multiply 6 by 11 to get 66, and then 6 by x to get 6x. Therefore, our answer is:
66 + 6x
Hope this helps!
Lady_Fox [76]3 years ago
4 0
6 times 11 + 6 times x
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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

Integrate by parts (twice).

\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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Solve 5a - 4a = -9 - 8a
Elena-2011 [213]
5a - 4a = -9 - 8a
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3 years ago
Read 2 more answers
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