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3 years ago

What is 4 copies of 1/5

Mathematics

1 answer:

GarryVolchara [31]3 years ago

6 0

Answer:

1/5 + 1/5 + 1/5 + 1/5 = 4/5

Step-by-step explanation:

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Question 5

Stells [14]

Answer:

7**6

Step-by-step explanation:

The expression 7^6 ; can be interpreted as 7 raised to the power of 6 ; which is (7 * 7 * 7 * 7 * 7 * 7). To execute this expression using a computer or basic mathematical operation on a computer we use the power symbol which the computer understand this the double multiplication symbol (**)

Hence, 7^6 = (7**6)

7 0

3 years ago

A cell phone company is looking to place a new tower to service 3 cities. In order for it to best service each city, it needs to

AlladinOne [14]

A circle is a collection of points all of which are on the circumference of the circle. If you use algebra and (x-h)^2 + (y-k)^2 = r^2 and the info contained in the coordinates (x,y) of each city, you can find h, k and r. Then r is the (equal) distance between each city and the center of the circle, where the new tower will be built.

3 0

3 years ago

Is every terminating decimal an integer?? Yes or no

sweet [91]

Integers do not have decimals...so ur answer is no. terminating decimals are not integers

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

Please help me if you help me the first one will get brainless

alexgriva [62]

Answer:

2/5

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers