Simplify using positive exponents only (2x^3y^3)^-2(4^-1x^5y^-2)^2

Select the ordered pairs that are solutions to the inequality 2x-3y greater than or equal to 12

Dima020 [189]

Just rank every pair of numbers for example:
The first one is: 1,-3.
2•1-3•-3>12
2+9>12
11>12
That's Incorrect. The first pair of numbers doesn't match the inequality.
Now do the same thing to all the other pairs of numbers.

3 0

3 years ago

Read 2 more answers

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

I need help RATS :) EHEHEHEHEHEHEEHEHEHEHEHEHEHEHEHEHEEHEHHEHEEHEHEH

galina1969 [7]

Answer:

Ok, so the answer is 3.5

Step-by-step explanation:

From what I could see it says walked for 14 around her block 4 times so your answer is 3.5 because if (Let's call her Patricia) Patricia walked around her block 4 times in 14 minutes, we have to find the unit rate, therefore we have to divide 14 by 4 as shown below

$14/4=3.5$

Thus your only viable solution is 3.5

Hope this Helps! Also, I would appreciate it if I could be rewarded with Brainliest, I work hard on these answers and I would enjoy it, you see my goal is to reach Genius status, as to provide many more helpful answers and help many more people. Even so, I hope that I have come of assistance to you!

4 0

2 years ago

Read 2 more answers

Can someone please help me with this ?

PtichkaEL [24]

Answer:

E. 2i√5

Step-by-step explanation:

√-20 = (√-1)(√20) = i·(2√5) = 2i√5

4 0

3 years ago

P1ss uyfuctfygui98hygcfxd hvbnnjbhgvfrdestdrygjh

lisov135 [29]

Answer:

the answer is 2 which is B.

6 0

3 years ago