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ikadub [295]
2 years ago
12

Help is need! look at the question

Mathematics
1 answer:
ser-zykov [4K]2 years ago
7 0

Answer:

Yes

Step-by-step explanation:

Plug in the numbers: 3≤3-2(0)

3≤3-0

3≤3

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Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.
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(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:

\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^{-x^2}\,\mathrm dx=1

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\displaystyle\frac12\int_0^\infty ce^{-u}\,\mathrm du=\frac c2\left(\lim_{u\to\infty}(-e^{-u})-(-1)\right)=1

which reduces to

<em>c</em> / 2 (0 + 1) = 1   →   <em>c</em> = 2

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\displaystyle\int_1^3 2xe^{-x^2}\,\mathrm dx = \boxed{\frac{e^8-1}{e^9}} \approx 0.3678

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2 years ago
Will give brainliest plz show work ASAP
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Step-by-step explanation:

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Nat2105 [25]

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Andrews [41]

Answer:

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Step-by-step explanation:

Let's make every fraction improper first.

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2 years ago
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