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Vedmedyk [2.9K]

2 years ago

14

Four internet providers offer similar services to almost everyone in the city. Any new company would have to engage in a price w

ar with the existing companies

1 answer:

aniked [119]2 years ago

6 0

What are you asking? If it’s true or false it’s TRUE?

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You buy shrimp at the market.

postnew [5]

Two 2.5 pound bags = 5 pounds for $34.99
34.99/5 pounds = 6.998 or $7/pound

8 0

2 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

When the absolute value of slope gets bigger the graph of the line gets ???

nekit [7.7K]

The slope of a line is defined as rise by run i.e.
how much y value changes with respect to the
change in × value. So the slope gives the
steepness of the line.
Slope O indicates there is no change in y when
the x value is changed and so the line with
slope 0 is a horizontal line. A line with slope 1
shows with 1 unit change in ×, y changes by 1
unit. Likewise slope equal 2 indicates with 1 unit
change in ×, y changes by 2 units.
We can see as the absolute value of slope is
increasing, the change in y with respect to x is
increasing i.e. the line is becoming more tilted or
steep.
Hence, we can conclude that:
The graph of a line gets Steeper as the absolute
value of the slope gets bigger.
C is the correct answer.

8 0

2 years ago

5+-6-45 plz help me I need it

kondor19780726 [428]

It is -46 that is the answer

4 0

3 years ago

Please help, pass due!

kondaur [170]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the parallel,lines

Since both the angles are corresponding angles, hence they are equal,so we get as,

7x+40 = 3x+112

===> 7x - 3x = 112-40

===> 4x = 72

===> x = 72/4 => 18

hence x = 18°

now angle A = 7(18)+40

A = 166°

hence, A = 166°

7 0

2 years ago