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3 years ago

Amelia has a 1-6 number cube (dice). She rolls the cube twice. What is the probability that she rolls two even

Mathematics

1 answer:

ryzh [129]3 years ago

4 0

Answer:

6/9

Step-by-step explanation:

2/6 x 2 = 0.6... and the 6 keeps repeating. There is one repeating number, so your answer is 6/9.

You might be interested in

The number of significant figures in 0.01500 is

madreJ [45]

Four significant is an answer
because if 0 is the extreme right of decimal place it is significant
and if 0 lies between two significant figure than it is significant like 1.020 is 4 sifnificant and 100 is 1 significant figure

8 0

3 years ago

Evaluate the function <br> f(8) + g(-1) for the functions f(x) = -2x + 9 and g(x) = x²-2

EastWind [94]

Answer:

- 8

Step-by-step explanation:

To evaluate f(8) substitute x = 8 into f(x)

f(8) = - 2(8) + 9 = - 16 + 9 = - 7

Similarly for g(- 1)

g(- 1) = (- 1)² - 2 = 1 - 2 = - 1

Then

f(8) + g(- 1) = - 7 + (- 1) = - 7 - 1 = - 8

7 0

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.

6 0

3 years ago

Expand. (1-4m)(m^2-3m+8) your answer should be a polynomial in standard form.

sergiy2304 [10]

Answer:

13m²−35m+8−4m³

Step-by-step explanation:

Expand (1 − 4m)(m² − 3m + 8) by multiplying each term in the first expression by each term in the second expression.

1m² + 1 (−3m) + 1 ⋅ 8 − 4m ⋅ m² − 4m (−3m) − 4m ⋅ 8

Simplify terms.

13m²−35m+8−4m³

8 0

2 years ago

How do you do that problem right there ?

jenyasd209 [6]

Haven't done this in awhile but i got
63x (2/9-6)
63x (0.2-6)
-364

6 0

3 years ago