All categories

2 years ago

Help pleasewhat is

Mathematics

1 answer:

RSB [31]2 years ago

8 0

Answer:

sure how can I help you? please tell me

You might be interested in

Using long division what is it a decimal equivalent of 11 I mean just a littl Using long division what is it a decimal equivalen

Vladimir [108]

Answer:

11.0 is the equivalent the decimal equiv of 9/1 is 9.0

Step-by-step explanation:

4 0

3 years ago

Hal is asked to write an exponential function to represent the value of a $10,000 investment decreasing at 2% annually. What mul

bogdanovich [222]

Hi there
The function is
Y (x)=10000 (1-0.02)^x
Y (x)=10000 (0.98)^x
Where x is the number of time

Hope it helps

4 0

3 years ago

Read 2 more answers

Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find

adoni [48]

a. I assume the following definitions for covariance and correlation:

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

$\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}$

Recall that

$E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

where $f_{X,Y}$ is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):

$E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$

$E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$

$E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2$

Also, recall that the variance of a random variable $X$ is defined by

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We use the previous fact to find the second moments:

$E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$

$E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$

Then the variances are

$\mathrm{Var}[X]=2-1^2=1$

$\mathrm{Var}[Y]=6-2^2=2$

Putting everything together, we find the covariance to be

$\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}$

and the correlation to be

$\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}$

b. To find the conditional expectations, first find the conditional densities. Recall that

$f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)$

where $f_{X\mid Y}$ is the conditional density of $X$ given $Y$ , and $f_X$ is the marginal density of $X$ .

The law of total probability gives us a way to obtain the marginal densities:

$f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$

Then it follows that the conditional densities are

$f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x$

$f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x$

Then the conditional expectations are

$E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}$

$E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}$

c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say

$E[X]=E[E[X\mid Y]]$

$\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]$

We've found that $E[X\mid Y]=\frac Y2$ and $E[Y\mid X]=X+1$ , so that by the LTE,

$E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X]$

$E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1$

$\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}$

Next, we have

$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}$

where the second moment is computed via

$E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$

In turn, this gives

$E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12$

$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12$

$\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}$

5 0

3 years ago

What is 48/100 reduced

garri49 [273]

Simplify: $\frac{48}{100}$
Both the numerator and the denominator are divisable by 2.
So:
$\frac{48~ divided ~by~ 2}{100 ~divided~ by~ 2}$ ≈ $\frac{24}{50}$

Still this fraction can be reduced lower, so divide by two again.

$\frac{24 ~divided~ by ~2}{50 ~divided~ by~ 2}$ ≈ $\frac{12}{25}$

Now, your final answer that is reduced (simplified) down to its lowest terms would be:

$\frac{12}{25}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-Hope this helped :)

7 0

3 years ago

Read 2 more answers

How do you divide 7 over 9

saw5 [17]

Answer:

The answer is 0.7 recurring.

Step-by-step explanation:

0.777.....

----------------------------

9 ) 7.000000000

70 and so on......

The 7 after the decimal point goes on without bounds.

7 0

3 years ago

Help pleasewhat is ​

Help pleasewhat is