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3 years ago

How many ways can a coach choose the 6 starting players from a volleyball team of13 players?

1 answer:

8 0

So slot meathod
6 slots, 13 in first
2nd=13-1 sice 1 used for first
and so on till 6 slots filled
so we have
13,12,11,10,9,8
mutily them
13 times 12 times 11 time s10 times 9 times 8=1235520 ways

You might be interested in

If the zeros are 1 and -4, what are the factors?

Natali5045456 [20]

Answer:

(x - 1) and (x + 4)

Step-by-step explanation:

Given the zeros of a polynomial, say x = a, x = b then the factors are

(x - a), (x - b) and the polynomial is the product of the factors, that is

f(x) = (x - a)(x - b)

Here the zeros are x = 1 and x = - 4, thus the factors are

(x - 1) and (x - (- 4)), that is (x - 1) and (x + 4)

5 0

3 years ago

PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!

pantera1 [17]

Answer:

N= -4

Step-by-step explanation:

Since mid-point of -13 and 3 is -4.

4 0

2 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.

6 0

3 years ago

A²- b² =<br> a² + b² =<br> (a + b)³ =

trasher [3.6K]

Answer:

a²- b²=(a+b)(a-b)

a² + b²=(a+b)²-2ab or (a-b)²+2ab

(a + b)³=a³+3a²b+3ab²+b³ or, a³+b³+3ab(a+b)

5 0

3 years ago

Five less than the quotient of x and three.

Darina [25.2K]

Answer:

(x/3) - 5

Step-by-step explanation:

Five less than quotient of x and three

7 0

3 years ago

Read 2 more answers