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3 years ago

How do you find the area of combined rectangles?

2 answers:

4 0

Find the area for each triangle with the formula being length (side) times width (base or bottom) times the height (this starts from the top point of the triangle and goes straight down to the base so it could be the middle or a side) After getting the area for each triangle, add them both together.

Vadim26 [7]3 years ago

3 0

Find the area of one rectangle and then the other and add the two together.

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Let Y1, Y2, . . . , Yn be independent, uniformly distributed random variables over the interval [0, θ]. Let Y(n) = max{Y1, Y2, .

Anettt [7]

Answer:

a) $F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

b) $f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

c) $E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

Step-by-step explanation:

We have a sample of $Y_1, Y_2,...,Y_n$ iid uniform on the interval $[0,\theta]$ and we want to find the cumulative distribution function.

Part a

For this case we can define the CDF for $Y_i$ , $i =1,2.,,,n$ like this:

$F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

Part b

For this case we know that:

$F_{Y_{(n)}} (y) = P(Y_{(n)} \leq y) = P(Y_1 \leq y,....,Y_n \leq y)$

And since are independent we have:

$F_{Y_{(n)}} (y) = P(Y_1 \leq y) * ....P(Y_n \leq y) = (\frac{y}{\theta})^n$

And then we can find the density function calculating the derivate from the last expression and we got:

$f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

Part c

For this case we can find the mean with the following integral:

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y y^{n-1} dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^n dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+1}}{n+1} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

For the variance first we need to find the second moment like this:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^2 y^{n-1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^{n+1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+2}}{n+2} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+2}}{n+2}= \theta^2 [\frac{n}{n+2}]$

And the variance is given by:

$Var(Y_{(n)}) = E(Y^2_{(n)}) - [E(Y_{(n)})]^2$

And if we replace we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2}] -\theta^2 [\frac{n}{n+1}]^2$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2} -(\frac{n}{n+1})^2]$

And after do some algebra we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

3 0

4 years ago

There are 4 cups of flour in 3 batches of cookies . How many cups are in one batche.

s344n2d4d5 [400]

Answer: Around 1.33 Cups per batch

Step-by-step explanation:

4 / 3 ≈ 1.33

4 0

3 years ago

Read 2 more answers

Help! What is m∠A ? ____°

beks73 [17]

I believe it is 72 because there is a right angle so thats 90 the known angle is 18 so thats 18-90-18=72

4 0

3 years ago

Using the function below, evaluate f(-2). * 2 points f(x) = 3x? - 5x - 2

monitta

Okay so I'm guessing that they are two different formulas the f(x)= 3x and the f(x)= -5x-2. If that's is true then you would just have to plug in -2 to the x's. So then it would be f(x)= 3(-2) which if you solve it would give you f(x)= -6 and for the second one it would be f(x)= -5(-2)-2 so then you would solve for it and get f(x)= 8.

6 0

3 years ago

Please help me with these two

Luda [366]

Answer:

Step-by-step explanation:

4 0

3 years ago