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Katyanochek1 [597]
3 years ago
15

The Mid-Atlantic Ridge is a spreading crack between crack between continental plates and it moves apart at 2.5 cm per year. New

York is 5567 km from London. In 1,000,000 years, how many far apart will New York be from London.
Mathematics
1 answer:
satela [25.4K]3 years ago
3 0
In 1,000,000 years, if the Mid-Atlantic Ridge continues to crack at a constant rate, New York and London will be 5592 km away from each other. I found this out by using the unit rate 2.5 cm per year to find out that in 1 million years it would be 2,500,000 cm. There are 100,000 cm in a kilometer. Therefore 2,500,000 cm is equivalent to 25 km. Add the 25 km to 5567 km to get 5592 km as the distance between the two cities.
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What is the solution to the system of equations? -2x + y = -5 3x – 2y = 12 A) (3, 1) B) (6, 3) C) (-2, -9) D) (-2, -1)
max2010maxim [7]

Answer:

The correct answer is D) (-2, -1)

Step-by-step explanation:

In order to solve this system of equations, start by multiplying the entire first equation by 2. Then add the two equations together. This will get the y's to cancel and allow you to solve for x.

-4x + 2y = -10

3x - 2y = 12

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

-x = 2

x = -2

Now that we have the value for x, we can find y by plugging the x value into either equation.

-2x + y = -5

-2(2) + y = -5

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3 years ago
If X and Y are independent continuous positive random
Leni [432]

a) Z=\frac XY has CDF

F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy

where the last equality follows from independence of X,Y. In terms of the distribution and density functions of X,Y, this is

F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy

Then the density is obtained by differentiating with respect to z,

f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy

b) Z=XY can be computed in the same way; it has CDF

F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Differentiating gives the associated PDF,

f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Assuming X\sim\mathrm{Exp}(\lambda_x) and Y\sim\mathrm{Exp}(\lambda_y), we have

f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}

and

f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

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