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2 years ago

PLEASE HELP!!! WILL GIVE BRAINLIEST AND ITS WORTH 30 points!!

Mathematics

2 answers:

Art [367]2 years ago

6 0

Since it’s a parallelogram, AB is parallel to DC and AD is parallel to BC. So, we can start looking at angle relationships. my drawing is horrible, but I basically drew a corresponding angle (congruent if lines are parallel) to DAB, which is then alternate interior angles with Y (congruent if lines are parallel) so, 145

olasank [31]2 years ago

3 0

145*.

Hope this helps

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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

2 years ago

Rachel traveled to five different areas (A, B, C, D, and E) to study the number of buckeye butterflies and the number of monarch

Ivenika [448]

Answer:

As we can see our our ration in Area B and Area D gives the same answer, so they are in proportion

Step-by-step explanation:

Given that

Rachel traveled to 5 areas to find number of Buckeye and Monarch butterflies.

Areas Rachel Traveled to = A, B, C, D, E

The table given below shows her findings.

Area Buckeye Monarch

A 15 16

B 27 36

C 12 25

D 24 32

E 44 33

Now lets find the ratio of Buckeye and Monarch Butterflies in these areas

Area A

Buckeye : Monarch = 15:16 = 0.9375

Area B

Buckeye : Monarch = 27:36 = 0.75

Area C

Buckeye : Monarch = 12:25= 0.48

Area D

Buckeye : Monarch = 24:32 = 0.75

Area E

Buckeye : Monarch = 44:33 = 1.334

As we know that if two ratios give same answer after solving the fraction then these ratios are in proportion.

As we can see our our ration in Area B and Area D gives the same answer, so they are in proportion.

4 0

2 years ago

Read 2 more answers

-7h+3(4-6h)=127 is h= 25?

GREYUIT [131]

Answer:

No, h = (-23)/5

Step-by-step explanation:

Solve for h:

3 (4 - 6 h) - 7 h = 127

3 (4 - 6 h) = 12 - 18 h:

12 - 18 h - 7 h = 127

-18 h - 7 h = -25 h:

-25 h + 12 = 127

Subtract 12 from both sides:

(12 - 12) - 25 h = 127 - 12

12 - 12 = 0:

-25 h = 127 - 12

127 - 12 = 115:

-25 h = 115

Divide both sides of -25 h = 115 by -25:

(-25 h)/(-25) = 115/(-25)

(-25)/(-25) = 1:

h = 115/(-25)

The gcd of 115 and -25 is 5, so 115/(-25) = (5×23)/(5 (-5)) = 5/5×23/(-5) = 23/(-5):

h = 23/(-5)

Multiply numerator and denominator of 23/(-5) by -1:

Answer: h = (-23)/5

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2 years ago

How does a change in the value of h effect the axis of symmetry of the graph ? How does the value of k change the axis of symmet

Ber [7]

The value of h is exactly the axis of symmetry. So when h changes, it completely changes the line of symmetry. This is because the vertex is (h, k) and the x value of the vertex is always equal to the line of symmetry in a quadratic.

Changing the k value does nothing to the line of symmetry. This is because it is just moving the graph up and down, which doesn't change the symmetry of a parabola.

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2 years ago

For the problem 135 divided by 5, draw two different ways to break apart the array.Use the Distributive Property to write produc

777dan777 [17]

The correct answer is 27

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2 years ago