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3 years ago

Pets owned by each customer or jays pet store

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

Step-by-step explanation:

its the median, just see which one is in the middle by crossing out one dot on each side until you get to the middle

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Y+3=20 simplify answer

ludmilkaskok [199]

Answer:

y=17

Step-by-step explanation:

subtract 3 from both sides.

4 0

3 years ago

Read 2 more answers

Can you please help with these questions? I need to finish homework by tonight.

noname [10]

Length x width is the formula.
so 18 x 18 for 7.
22 x 24 for 8.
and 8 would probably be 5 x 5 to equal 20. so the length and width would be 5 and 5 because 20 is the area

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

PLS HELP WILL MARK YOU BRAINLIEST, NO FAKE ANSWERS!!

Makovka662 [10]

Answer:

1. if x + 5 = 12, then x = 7

:Subtraction property of equality

2. If x + y = 20, and y = 5, then x + 5 = 20.

<u>:Substitution property of equality</u>

3. If 2x3 = 11, then 11 = 2x - 3.

<u>:Symmetric property of equality</u>

3 0

3 years ago

A group of 29 friends get together to play a sport first person must be divided into team each team has to have exactly 4 player

telo118 [61]

Answer:

7 teams remaining a person

Step-by-step explanation:

29/4 = 7 remainder 1

5 0

2 years ago