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3 years ago

HELP ASAP PLEASE!!!!

Mathematics

2 answers:

Margaret [11]3 years ago

8 0

Answer:

answer is 16 degree x=16

riadik2000 [5.3K]3 years ago

7 0

It is 16 because you subtract 180 and 133 which is 47 then add 117 which is164 then subtract 180-164 which is 16

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Help me plz right answer

boyakko [2]

What’s up again douglas. it would be B

4 0

3 years ago

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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

HELP ME AS FAST AS YOU CAN!

Kryger [21]

Answer:

Step-by-step explanation:

2x. add 8 to other side. you get 2x=12

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3 years ago

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In 3 hours a car traveled 180 kilometers. At the same average rate, how many kilometers can the car travel in 5 hours?

arlik [135]

Answer: 5 hours = 300 Kilometers

Step-by-step explanation: If a car can travel 180 kilometers in three hours then in one and a half hour that car can travel 90 kilometers. If you divide 90 by 3 you will find out that in a half hour that car can travel 30 kilometers. So add 180+90+30=300 that is the same as adding three hours+one and a half+thirty minute=five hours

Hope This Helps!!!!!

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4 years ago

The segments shown below could form a triangle.<br> O<br> A. True<br> O<br> B. False

andrew11 [14]

Answer:

What segment shown below!?!?!?!?!

Step-by-step explanation: I DON'T SEE ANY SEGMENT SHOWN BELOW!!! PLZ HELP!

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3 years ago

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