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1 year ago

4. Lucy is k years old. She has a sister who is three years younger than she is and

Mathematics

1 answer:

WITCHER [35]1 year ago

7 0

Answer:

4k - 8

Step-by-step explanation:

Lucy is k years old so we know k is an important part of this problem.
Sister 1 is 3 years younger, so she is k - 3 years old.
Sister 2 is 5 years less than twice Lucy's age therefore, she is 2k (twice Lucy's age) - 5 years old.
Now we have: k + (k - 3) + (2k - 5)
We can simplify like terms with the k's and get 4k. then we add (-3) and (-5) to get 4k - 8.

Hope this helps!

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Kim eats 1/3 of an apple in 1/6 minute.How many minutes will it take

klio [65]

Answer:

0.5 or 3/6 min

Step-by-step explanation:

1/3= 1/6min or 10 sec

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1 part is 10 sec, 3 parts will be 30 secs which is 0.5 /3/6 min.

hope this helps

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2 years ago

Please help me thanks

drek231 [11]

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2 years ago

Write the explicit formula that represents the geometric sequence-2, 8, -32, 128

maks197457 [2]

Well, since we know is a geometric sequence, we can always get the common ratio of it by simply dividing one value by the one behind it... so let's do so, with say hmm -32 and 8 -32/8 = -4 <-- our common ratio

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$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
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n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=-2\\
r=-4
\end{cases}\implies a_n=-2(-4)^{n-1}$

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2 years ago

Find the value of x and y.<br><br><br>I will give brainliest to the best awnser, thank you.

frez [133]

Answer:

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Step-by-step explanation:

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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

2 years ago