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3 years ago

What is equivalent to (3 root 8^1/4 x) ?

Mathematics

1 answer:

vfiekz [6]3 years ago

7 0

Answer:

The given expression is equivalent to 3(2^(3/8))x.

Step-by-step explanation:

The given expression (3 root 8^1/4 x) can be written as:

3√(8^(1/4))x

we know that 2^3 = 8

= 3√(2^(3/4))x

We know that √x = (x)^(1/2)

= 3(2^((3/4)(1/2))x

Powers multiply with each other, 3*1/4*2 = 3/8

= 3(2^(3/8))x

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Angelina_Jolie [31]

Answer:

1) 3

2)9

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4)18k

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3 years ago

The area of a rectangle garden is 2g^2+34g+140. If the width is 2g+14 what is the length?

ASHA 777 [7]

We know that
[area of rectangle]=length*width
area=2g²<span>+34g+140
</span><span>width =2g+14
</span>
step 1
find the roots of 2g²+34g+140

using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
g=-10
g=-7
so
2g²+34g+140=(2g+14)*(g+10)

therefore

the length is (g+10)

the answer is
the length is (g+10)

3 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

What number is needed to complete the pattern. 66 73 13 21 52 __ 10 20

svetoff [14.1K]

Please try 43 I'm not sure

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3 years ago

Solve the system of equations<br> -5x+13y+-7<br> 5x+4y=24<br> y=<br> x=

aleksklad [387]

<h3>The value of y is equal to 1.</h3><h3>The value of x is equal to 4.</h3>

Because both equations have a term that will cancel out if they're added together, we're going to add both equations together.

-5x + 5x = 0

13y + 4y = 17y

-7 + 24 = 17

17y = 17

Divide both sides by 17.

y = 1

Now that we have a constant value of y, we can solve for x.

-5x + 13(1) = -7

-5x + 13 = -7

Subtract 13 from both sides.

-5x = -20

Divide both sides by -5.

x = 4

3 0

3 years ago

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