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My name is Ann [436]

4 years ago

11

How do I solve −2(n− 9) + 4 using distributive property or combining like terms?

1 answer:

ryzh [129]4 years ago

8 0

Answer:

Multiply the term outside of the parentheses by each term in the parentheses. To do this, you are essentially distributing the outer term into the inner terms. Multiply the term outside the parentheses by the first term in the parentheses. Then multiply it by the second term. If there are more than two terms, keep distributing the term until there are no terms left. Keep whatever operation (plus or minus) is in the parentheses

Step-by-step explanation:

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Suppose that Y1, Y2,..., Yn denote a random sample of size n from a Poisson distribution with mean λ. Consider λˆ 1 = (Y1 + Y2)/

Burka [1]

Answer:

The answer is " $\bold{\frac{2}{n}}$ ".

Step-by-step explanation:

considering $Y_1, Y_2,........, Y_n$ signify a random Poisson distribution of the sample size of n which means is λ.

$E(Y_i)= \lambda \ \ \ \ \ and \ \ \ \ \ Var(Y_i)= \lambda$

Let assume that,

$\hat \lambda_i = \frac{Y_1+Y_2}{2}$

multiply the above value by Var on both sides:

$Var (\hat \lambda_1 )= Var(\frac{Y_1+Y_2}{2} )$

$=\frac{1}{4}(Var (Y_1)+Var (Y_2))\\\\=\frac{1}{4}(\lambda+\lambda)\\\\=\frac{1}{4}( 2\lambda)\\\\=\frac{\lambda}{2}\\$

now consider $\hat \lambda_2$ = $\bar Y$

$Var (\hat \lambda_2 )= Var(\bar Y )$

$=Var \{ \frac{\sum Y_i}{n}\}$

$=\frac{1}{n^2}\{\sum_{i}^{}Var(Y_i)\}\\\\=\frac{1}{n^2}\{ n \lambda \}\\\\=\frac{\lambda }{n}\\$

For calculating the efficiency divides the $\hat \lambda_1 \ \ \ and \ \ \ \hat \lambda_2$ value:

Formula:

$\bold{Efficiency = \frac{Var(\lambda_2)}{Var(\lambda_1)}}$

$=\frac{\frac{\lambda}{n}}{\frac{\lambda}{2}}\\\\= \frac{\lambda}{n} \times \frac {2} {\lambda}\\\\ \boxed{= \frac{2}{n}}$

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3 years ago

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Y=-3x+9 that is the answer

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3 years ago

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Then:

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There are 12 blue hats.

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2 years ago