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Alinara [238K]
3 years ago
5

A group of 10 people is choosing a chairperson and vice-chairperson. They put all 10 people's names into a hat. The first name d

rawn becomes chair. The second name drawn becomes vice-chair. How many possible combinations of chair and vice-chair are there?
Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

10

Step-by-step explanation:

A group of 10 people is choosing a chairperson and vice-chairperson. They put all 10 people's names into a hat. The first name drawn becomes chair. The second name drawn becomes vice-chair. How many possible combinations of chair and vice-chair are there?

19

90

100

10! (10 factorial)

Youre Welcome! :)

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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}

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Tom and Lana were comparing their Halloween candy. Tom received 4 times as much candy as Lana received. Tom then split his candy
seropon [69]
Given that Lana received 18 ounces of candy and Tom received 4 times as much candy as Lana received, then the number of candy Tom received is given by 4(18) = 72.

Now, given that Tom divided his candy evenly into 9 piles. Then the number of ounces of candy in each of Tom's piles is given by:

72/9 = 8 ounces.
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Mr. Curry puts $5,500 in an investment account that offers 8% interest compounded annually. He makes no additional deposits or w
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Answer:

B.$12,100.00

Step-by-step explanation:

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What fraction of a pound is 25 pence
scZoUnD [109]
There are 100 pence in a Pound. So  we have to divide 25 pence over this pound, or over 100 pence: 25/100 is 1/4, so 25 Pence is one forth of a Pound (or 25 % in percentages): 
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