3 years ago

20 POINTS ANSWER FAST

Mathematics

2 answers:

Monica [59]3 years ago

8 0

I’m pretty sure the answer is (2,1)

Gekata [30.6K]3 years ago

6 0

Answer:

My opinion for the answer of this question would be (2,1)

Step-by-step explanation:

Please give me brainliest :)

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Establish for which constants c the following function is a density.

Valentin [98]

By "density" I assume you mean "probability density function". For this to be the case for $f(x)$ , we require

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

Since

$f(x)=\begin{cases}cx^{1/2}&\text{for }0$

you have

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_0^2cx^{1/2}\,\mathrm dx=\dfrac{2c}3x^{3/2}\bigg|_{x=0}^{x=2}=\dfrac{2^{5/2}c}3=1$

which means

$c=\dfrac3{2^{5/2}}=\dfrac3{4\sqrt2}$