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Marina CMI [18]
3 years ago
15

**Jane found money in her pocket. She went to a convenience store and spent 1/4 of her money on chocolate milk, 3/5 of her money

on a magazine, and the rest of her money on candy. What fraction of her money did she spend on candy? ​
Mathematics
2 answers:
valentina_108 [34]3 years ago
8 0

Answer:

3/20

Step-by-step explanation:

1/1-1/4-3/5= (money spent)

1×4×5/(1×4×5)-1×1×5/(1×4×5)-3×1×4/(1×4×5)

20/20-5/20-12/20

(20-5-12)/20=3/20

Oksi-84 [34.3K]3 years ago
5 0

Answer:

$y - $0.85

Step-by-step explanation:

y represents how much money he had.

\frac{3}{5}+\frac{1}{4} =\frac{17}{20}

$y - $0.85

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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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