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3 years ago

What prperty did you use

Mathematics

1 answer:

nekit [7.7K]3 years ago

3 0

The property is distributive

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Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

5 0

3 years ago

Nancy created a graph to predict the pay she will take home after taxes are taken from her income. Using the information on the

deff fn [24]

The answer is c. 0.8

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3 years ago

Read 2 more answers

3x square+6y Square when X=0 and y=2 with steps pls

Sauron [17]

Answer:

If the equation is 3x^2+6y^2, when x=0 and y=2.

Then, 3(0)^2+6(2)^2=

So, 0+6(4)= 24

Therefore, the answer is 24.

Step-by-step explanation:

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3 years ago

If a radius of a circle is perpendicular to a chord, then it _____ that chord.

labwork [276]

The answer is b. There is a proof, but it is easily got by proving the other cases wrong.

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3 years ago

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List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers,

Anarel [89]

Answer: YES

Step-by-step explanation:

We need to write out the expressions

P= {m}

Q= {n}

R= {m+n}

If 2m=n then we can say;

P= {½n} Q= {n} & R= {³/²n}

It is obvious that the smaller number in Q is greater than the largest number in P

We can make some assumptions.

Let n= (x,y,z)

Consequently,

P={½x,½y,½z} Q={x,y,z} and R= {1.5x,1.5y,1.5z}

Therefore the median will be the middle element,

Median of P= ½y

Median of Q = y

Median of R = 1.5y

And 1.5y>1.5y

Then we can agree that the median of R is greater than the median of both P and Q

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4 years ago