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3 years ago

Snnsnsnsnnsnsnsnsnsnsn

Mathematics

1 answer:

Alina [70]3 years ago

7 0

Answer:

x=1+√17 or x=1−√17

Find Discriminant

Step-by-step explanation:

Brainliest pleaseeeeeeeeeeeeeeeeeeeeeee

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HELP NEEDED!

Ilia_Sergeevich [38]

Answer:

x = 12.6

Step-by-step explanation:

Sum of measure of all the arcs of a circle = 360°

m(arc CD) + m(arc CED) = 360°

Since, m(arc CED) = (20x)°

And m(arc CD) = (8x + 6)°

By substituting these measures in teh expression,

(20x)° + (8x + 6)° = 360°

(20x + 8x) + 6 = 360

28x = 360 - 6

28x = 354

x = $\frac{354}{28}$

x = 12.64

x = 12.6

4 0

3 years ago

The table shows a pattern of exponents.

ahrayia [7]

Answer: c. divide the previous value by 5

Step-by-step explanation:

7 0

3 years ago

Find the h.c.f by prime factorization method 27, 30

Orlov [11]

Answer:

Find it yourself . you must try yourself. rate me plzs

7 0

3 years ago

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :