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3 years ago

Is this right? I feel like I’m wrong.

Mathematics

2 answers:

Llana [10]3 years ago

5 0

Answer:

That is completely correct!

Arisa [49]3 years ago

4 0

Answer:

no I believe you are correct

Hope This Helps! Have A Nice Day!!

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If you had a billion dollars and you spend one dollar every second how many years would it take you to spend it all

trapecia [35]

Answer: 31.71 years

Step-by-step explanation:

First, we need to find how many seconds are in a day.

There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.

There are 3,600 seconds in an hour. How did we find this? 60 (min) x 60 (sec)

There are 86,400 seconds in a day. 24 (hr) x 3,600 (sec)

Now we need to find how many seconds are in a year.

There are 365 days in a year, we can multiply that by 86,400.

365 x 86,400 = 31,536,000

Whew! Now that we have that out the way, we can now divide $1 billion by 31,536,000 seconds.

1,000,000,000/31,536,000 = 31.71 (rounded to the nearest hundredth)

If you had a billion dollars, it will take 31.71 years to spend one dollar every second.

3 0

3 years ago

Read 2 more answers

Which rule describes the translation?

max2010maxim [7]

Answer:

Step-by-step explanation:

translations assignment

4 0

3 years ago

8. Which equation best represents the line

jeka94

Step-by-step explanation:

first one

when x=0, y=4

wheb x=4, y=0

3 0

3 years ago

What does the x represent?<br><br> a.diameter <br> b. central angle<br> c. arc

IrinaVladis [17]

A because it shows hi wide the circle is⭕️

8 0

3 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

3 years ago