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3 years ago

What is the value of x? 63, 42, 50

Mathematics

1 answer:

taurus [48]3 years ago

6 0

Answer:

Step-by-step explanation:

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Can someone please help will mark brainlist

Virty [35]

Answer:

You can complete the mission 5 times before you earn less than 400 points.

Explanation:

$100(0.8)^n^-^1 < 400$

$(0.8)^n^-^1 < 0.4$

$log 0.8^n < log 0.32$

$n > 5.1$

8 0

2 years ago

Two trains are moving towards each other. One train moves at a speed of 50 mph, and the other train travels at 60 mph. How soon

Margaret [11]

I think that they will meet in 3 hours and 11 min but I’m not too sure

4 0

3 years ago

Read 2 more answers

Can anyone help me on this please?

rjkz [21]

$\dfrac{4a^3b^2+2a^2b^2+3ab}{ab}=\dfrac{4a^3b^2}{ab}+\dfrac{2a^2b^2}{ab}+\dfrac{3ab}{ab}=4a^2b+2ab+3$

8 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 $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago

Complete the description of the translation

Anit [1.1K]

The triangle translated 3 units right and 4 units down. Hope this helps!

4 0

3 years ago