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3 years ago

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Simplify the expression. (–2)5 –32 16 32 –10

2 answers:

wlad13 [49]3 years ago

5 0

Answer:

( 60 )

− 4

( − 45 )

( 390 )

e-lub [12.9K]3 years ago

3 0

-10 i hope that 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:

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

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3 years ago

Elle went to pet smart and bought 4 gold fish and three turtles for $28. later the day, warren went to Petsmart and bought six g

AlekseyPX

It would cost 40$ because you are doing 10 times 4

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3 years ago

Read 2 more answers

The sum of two numbers is 62, and their difference is 8. What are the<br> numbers?

Marina CMI [18]

Answer: 35 and 27

Step-by-step explanation:

We can solve this by using a system of equations

x will be the first number while y will be the second

$x+y=62$

This equation means the two numbers will add to get 62

$x-y=8$

This equation means when subtracted the two numbers will have the difference of 8.

Now I will solve by elimination

$x+y=62\\\\+x-y=8\\2x=70$

When adding the two equations why was eliminated so now we can find the value of x

$2x=70\\\frac{2x}{2} =\frac{70}{2} \\x=35$

Now sub in x into one of the original equations to find the value of y

$35+y=62\\35-35+y=62-35\\y=27$

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2 years ago

Solve equation by using the quadratic formula.

pishuonlain [190]

C. x=-2/3, -4/5 (negative two-thirds, negative four-fifths)

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2 years ago

Which number is not equivalent to 467%?

Burka [1]

It could be almost anything, but an example is three (3).

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3 years ago