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Alja [10]
4 years ago
12

What is 7/20 in decimals

Mathematics
2 answers:
FromTheMoon [43]4 years ago
6 0

Answer:

0.35 :)

Step-by-step explanation:

Musya8 [376]4 years ago
4 0

Answer:  0.35 your welcome

Step-by-step explanation:  

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A line has a slope of -2 and passes through the point (1,-2). what is an equation of this line ?
Ket [755]
Y=-2x would be the answer.
4 0
3 years ago
Read 2 more answers
Solve for the following 6(16x+3)-2/3(9-12x)+25=0
lutik1710 [3]
First expand the parentheses:-

96x + 18 - 6 + 8x + 25 = 0

104x = -18 + 6 - 25 =  -37

x = -37 / 104 =  -0.3558 
6 0
3 years ago
When x=8 and y=20 find y when x=42
ella [17]
When x=42 that means it is 8x6 so that means that the y = 120 becuase its 20x6
answer= y=120
7 0
4 years ago
Clark and his brother kent played floor hockey in their living room. Each goal was worth 2 points. Clark scored 6 points. Kent s
Schach [20]

Answer:

7 Goals

Step-by-step explanation:

6 divided by 2 equals 3, So Clark scored 3 goals. 8 divided by 2 equals 4, So Kent scored 4 goals. 4 plus 3 equals 7.

5 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
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