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Marysya12 [62]
3 years ago
10

A ski patrol unit has nine members available for duty, and two of them are to be sent to rescue an injured skier. In how many wa

ys can two of these nine members be selected? Enter you answer; In how many ways can two of these nine members be selected?
Mathematics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

36

Step-by-step explanation:

The number of two of the nine members of ski patrol unit can be chosen is equal to the combinations of 9 taken 2. This can be calculated using the formula:

\frac{9!}{2!*(9-2)!}  

= \frac{9!}{2!*7!}, where

9! =9×8×7×6×5×4×3×2×1 and

7!=7×6×5×4×3×2×1

2!=2×1   then the equation becomes:

=\frac{9*8}{2} =36

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The sum of two numbers is 27. One of the numbers is three more than the other. Find the numbers,
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Answer:

12 and 15

Step-by-step explanation:

x + y = 27

x +3 = y

x = y - 3

y - 3 + y = 27

2y = 24

y = 12

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Write the equation of a square root function that has the following graph.
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Answer:

  g(x) = √(x -5) +7

Step-by-step explanation:

To translate the graph of f(x) by h units horizontally and k units vertically, the function is transformed to ...

  g(x) = f(x -h) +k

To translate f(x) = √x by 5 units horizontally and 7 units vertically, the function is transformed to ...

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4i+4-2i+10<br> whats the answer to this equation 0_0
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2i + 14

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
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.

6 0
3 years ago
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