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2 years ago

Solve the equation 4 1/5+b= 9 3/5

2 answers:

8 0

4 1/5 + b = 9 3/5

4 1/5 - 4 1/5 + b = 9 3/5 - 4 1/5
(taking the same amount from each side to keep it balanced)

0 + b = 5 2/5
(subtracting 4 1/5 from each side)

b = 5 2/5

Alik [6]2 years ago

4 0

4 1/5+b=9 3/5
convert to improper fraction
21/5+b=48/5
since they both have the same denominators,you can subtract the numerators
b= 48/5-21/5
b=27/5= 5 2/5

hope this helps

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Allison’s dog is 6 pounds more than 5 times the weight of Gail’s dog. What algebraic expression could you make to find (g) which

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Given that Allison’s dog is 6 pounds more than 5 times the weight of Gail’s dog, it is important to know the following:

1. The word "more" indicates an Addition.

2. The word "times" indicates Multiplication.

Knowing this, you can write the following expression to represent "5 times the weight of Gail’s dog", where "g" is the weight of Gail’s dog:

$5g$

Therefore, you can determine that the weight of Allison's dog (in pounds) is the sum of 6 pounds and 5 times the weight of Gail's dog (in pounds). Then, you can represent this with this algebraic expression:

$5g+6$

Hence, the answer is:

$5g+6$

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5 months ago

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Harlamova29_29 [7]

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2 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.

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valentina_108 [34]

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Step-by-step explanation:

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emmainna [20.7K]

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Step-by-step explanation:

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