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2 years ago

If you give correct answer with the Explanation I mark you brainliest answer and give 5 stars

Mathematics

1 answer:

Sedaia [141]2 years ago

7 0

Answer:

x = +4√5, x = -4√5

Step-by-step explanation:

Given:

x² - 80

Find:

Zeroes

Computation:

x² - 80

x² - 80 = 0

x² = 80

x = √80

x = ±4√5

x = +4√5, x = -4√5

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A candle burned at a steady rate. After 37 minutes, the candle was 11.2 inches tall. Eighteen minutes later,

irinina [24]

Answer:

i think the answer is 4.65 inches tall.

Step-by-step explanation:

i attempted it this may be wrong just wait for someone to confirm it.

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3 years ago

the length of a rectangle is 5 cm less than 3 time its width. if the perimeter of the rectangle is 54 cm, find the length and wi

castortr0y [4]

$length = 3x - 5 \\ width \: = x \\ p = 2l + 2w \\ 54 = 2(3x - 5) + 2x \\ 54 = 6x - 10 + 2x \\ 54 = 8x - 10 \\ 64 = 8x \\ x = 8 = width \\ 3(8) - 5 = 24 - 5 = 19 \\ length = 19$

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2 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

I don't get this!!!!!Help please!!!!!!

Westkost [7]

I want to say the first one, but i do not know for sure.

8 0

2 years ago

-2x-9y=-25 -4x-9y=-23

goldfiish [28.3K]

So to start this off you would add 9y to both sides of the equation and then you regroup it which gives you −2x=−25+9y and then you regroup the terms and get -2x=9y-25 and then you dived both sides by -2 and get x=9y-25/-2 then you simplify and so on

but at the end you should get x=-1 and y =3

8 0

2 years ago