48+49+50+51+...112+113

The graphs below have the same shape. What is the equation of the red

Citrus2011 [14]

Answer:

g(x)=2-x² OR g(x)=-x²+2

Step-by-step explanation:

Notice that the red parabola doesn't stretch or compress. It only translates 3 units down. This means taking the original equation f(x)=-x²+5 and subtracting 3 units will get us the equation g(x)=-x²+2 which is also the same thing as g(x)=2-x². Looks like some of your answer choices got cut out of the picture so make sure to choose the answer I provided.

4 0

3 years ago

Read 2 more answers

Name the parallel segment

ratelena [41]

MN would be the answer to your question because they are parallel

4 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

How are the formulas for the area of a rectangle and the area of a triangle alike and different? And wolflover14, it's Nini on a

RoseWind [281]

Answer:

The formula to find the area of a rectangle is:

b * h = a

where b = base, h = height, and a = area.

The formula to find the area of a triangle is:

(1/2) * b * h = a

where b = base, h = height, and a = area.

They are similar because in both of the formulas, you are multiplying the base by the height. The difference is that in the triangle formula, you have to do an additional step of multiplying by (1/2) to get the area. hope this helps:)

4 0

3 years ago

Write a point-slope equation for the line that has slope 4 and passes through the point (11,7)

Sati [7]

Answer:

y-7=4(x-11)

Step-by-step explanation:

y-y1=m(x-x1)

y-7=4(x-11)

5 0

3 years ago