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Dmitry_Shevchenko [17]

3 years ago

8

Elliot got some new tradeing cards.He has 4 packs with 20 cards each.Another pack has only 8 cards.Whats another way to find how

many cards elliot has in total

2 answers:

swat323 years ago

8 0

Answer:

(4x20)+8=88

could be used to find out how many cards he has in total.

solniwko [45]3 years ago

6 0

Answer:

(4x20)+8=88

Step-by-step explanation:

thats the answer

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MariettaO [177]

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It is $4^{5}$

Step-by-step explanation:

the four is multiplied 5 times

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Can somebody tell me how to solve these kind of problems?

Ugo [173]

Answer: 918km^2

Step-by-step explanation:

$area(a): 1700km^2\\percentage(p): 5.75 = \frac{5.75}{100}=0.0575\\years(y): 8$

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$x=a-(a*p*y)$

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3 years ago

I’m getting ahead easily but I’m stuck with this question.

Kruka [31]

Answer:

I know #1 but not #2 it's

(10x2)+4=24 it's still less then 25 :)

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3 years ago

Can someone please help me with this question??

posledela

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In order of least money saved to most money saved: Katrina, Susan, Gabrielle, Savanna

Step-by-step explanation:

You can first change all of them to either percentages or fractions with the same denominator. I chose to use percentages. Thus Savanna's percentage of money saved would be 90%, Katrina would've saved 80% and Susan would've saved around 83%.

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3 years ago

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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