All categories

3 years ago

BRAINLIEST ANSWER TO THE FIRST CORRECT ANSWER!! PLEASE HELP!! ANSWER CORRECTLY PLEASE!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Nitella [24]3 years ago

5 0

The sequence is ×2.

12,288 is the 12th term.

24,570 is the answer. It is the sum.

You might be interested in

Which pairs of angles are supplementary? A) ∠BCD and ∠BAD B) ∠BCD and ∠ABC C) ∠BCD and ∠CDA D) ∠ABC and ∠BAD

Bogdan [553]

<span>∠ABC ∠BAD is the correct answer

</span>

4 0

3 years ago

there are 2 1\2 bus loads of students standing in a parking lot. The students are getting ready to go on a field trip. 2\5 of th

Archy [21]

It would take only one full bus to carry only girls.

3 0

3 years ago

Read 2 more answers

In the standard (x, y) coordinate plane, if the x-coordinate of each point is 2 times the y-coordinate, what is the slope of the

uranmaximum [27]

1/2 (Rise over Run)
I hope this helps out :3

3 0

3 years ago

Ricardo earns a base salary of $500 a week plus a weekly commission he has earned a commission of $325, $460, $280, $400, $380,

Vinvika [58]

Answers:

Mean = $349.2

Median = $352.5

Range = $210

$499.90

$355

I have attached a photo of the work written for this question:

4 0

4 years ago

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

Viefleur [7K]

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

b) For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

Step-by-step explanation:

For this case we assume that we have a random sample given by: $X_1, X_2,....,X_7$ and each $X_i \sim N (\mu, \sigma)$

Part a

In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

5 0

4 years ago