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Slav-nsk [51]
3 years ago
10

What is the slope of the line that connects the points (1,3)and (2,2)

Mathematics
1 answer:
Anna11 [10]3 years ago
8 0
You subtract the x intercepts so 1-2= -1 then you do the y intercepts so 2-3=-1 so -1/-1 would equal 1 because the negatives cancel out and anything over the same number would be 1
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Circle O is shown below. the diagram is not drawn to scale. 73°
zepelin [54]
Answer:
53.5

Explanation:
Since triangle AOB is an isosceles triangle (AO = OB), the base angles of the triangle are 53.5. (180-73 = 107, 107/2 = 53.5)
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The complement of a right angle is an obtuse angle. true or false.
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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}

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3 years ago
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Romashka-Z-Leto [24]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Solve the inequality and enter the solution as an inequality <br><br><br> X-7&lt;1​
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Answer: x < 8

Step-by-step explanation:

5 0
2 years ago
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