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2 years ago

What is a square a parallelogram but a parallelogram not a square a square

Mathematics

2 answers:

Savatey [412]2 years ago

5 0

Square is 90 degrees but parallelogram is not

Alinara [238K]2 years ago

3 0

A square has both parallel sides
being the top and bottom
but also both the sides

AND so does a PARALLELogram
BUT a square has ALL right (90 d) angles
and a parallelogram does NOT

ALSO . . . . . SUMMARY . . . . .
The sides of a square are parallel
and all the angles equal 360 degrees,
so it's a parallelogram. BUT a
parallelogram is NOT a square
because a squares angles need to be 90 degrees.

does that answer your question?

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Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

3 years ago

Someone please tell me and maybe ill pay you

rjkz [21]

4x = 6x -10
2x = 10
x = 5

answer
x = 5

6 0

3 years ago

If an object is projected upward from ground level with an initial velocity of 128 ft per sec, then it's height in feet after t

antoniya [11.8K]

Answer: t= 4 seconds

Maximum height = 256 feet

Step-by-step explanation:

The height of the object projected upwards after t seconds is given by

s(t)=-16t^2+128t.

The expression is a quadratic equation. When this equation is plotted on a graph, height against time, it takes the shape of a parabola whose vertex represents the maximum height attained by the object.

To get the value of t at the maximum height,

t = -b/2a

From the equation,

a = -16

b = 128

t = -128/-2×-16

= -128 /- 32= 4

it will take 4 seconds to reach its maximum height

To find maximum height attained, put t= 4 in the equation

s(t)=-16t^2+128t

S = -16× 4^2 + 128×4

S = -256+512

S = 256 feet

3 0

3 years ago

You walk at a speed of 1.429m/s. How far could you walk in 10 MINUTES? Show your work!

arsen [322]

Answer:

Well the answer is 857.4 meters per minute

Step-by-step explanation:

The reason is because 1.429 times 60= 85.74 That means that you walk 85.74 meters per second. Then that times 10 is 857.4 . Can I get brainliest

3 0

3 years ago

Cheri's time in the bobsled race was 1 min, 38.29sec. Write the word form and the value of the 9 in Cheri's time

Artemon [7]

I believe it would be the hundredth place

7 0

3 years ago