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GuDViN [60]
3 years ago
9

I could use any help as soon as possible in need this due tomorrow for exams

Mathematics
2 answers:
EastWind [94]3 years ago
6 0
The answer woukd be 23
nexus9112 [7]3 years ago
5 0
GCF for 27 and 14 is none
You might be interested in
Which value is closets to 8 11/20 A)8 B)8 1/2 C)8 3/4 D) 9
Alona [7]
I would say the answer is B. 8 1/2 because 11/20 is close to 10/20 which would be the same as 1/2
4 0
3 years ago
An athlete completes a race in 55.72 seconds. How many times greater is the digit in the tens place than the digit in the ones p
MArishka [77]
As they are the same the answer is 1 times
8 0
3 years ago
Which fraction does not belong with the other three explain 1/12 2/12 3/12 or 4/12
MakcuM [25]
1/12 does not belong with the other three fractions. 

2/12, 3/12, and 4/12 can all be simplified to 1/6, 1/4/, and 1/3. However, 1/12 can't be simplified.
6 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 nth 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
The graph shown compares the rent Andrew and Dave pay for renting bikes from different stores
Dmitry_Shevchenko [17]

Answer:I don’t know either

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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