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3 years ago

How do you write 0.04×10 to the 5th power in standard form?

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

4 0

0.04x10^5
move decimal 5 TIMES TO THE RIGHT.
0.04000
12345
4000
I hope this helps;)

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ZEARN MATH

joja [24]

Answer:

ok ill do it :-)

Step-by-step explanation:

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3 years ago

What is the cost of one cranberry and one banana nut muffin?

Bezzdna [24]

So for the cranberry muffin, for every 12 (a dozen) it's $3
Banana nut muffin: for every 12, it's $4.32
take the $3 and divide it by 12 which is $0.25 for every cranberry muffin
take $4.32 divide by 12 and you get $0.36 for every banana nut muffin
so $0.25 + $0.36 = $0.61
Therefore the answer is B

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3 years ago

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17. Which of the following is a graph of y= x^2?

Juli2301 [7.4K]

Answer:

The parabola - quadratic.

Step-by-step explanation:

There is only graph / option in the picture but it is the correct one. Regardless, whether x is a positive or negative, the y coordinate is going to be a positive number as x^2 will produce a positive. So as -x^2 will also be positive.

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2 years ago

the base of the Eiffel tower has a square with an area of 15625 square feet what is the length of the side of the base

tigry1 [53]

77374747484847373848484848

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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:

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

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3 years ago