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3 years ago

Plz help me solve this question algebraic expression 64p^4+81q^4

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer:

you can see in above photo

Step-by-step explanation:

hope it helps!!

Neporo4naja [7]3 years ago

3 0

Answer:

(8p+9q)⋅(8p−9q)

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Find the area of a regular hexagon with a base of 10 and an apothem of 20.

GarryVolchara [31]

I hope this helps you

regular hexagon =6 equilateral triangle

Area=6.(10.20/2)

Area=600

8 0

3 years ago

Read 2 more answers

I need help with this plzzz percent equations blank % of $900,000 = $270,000

notsponge [240]

Answer:

30%

Step-by-step explanation:

7 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

A dartboard has a diameter of 18 inches.What are its radius and circumference?

allsm [11]

Diameter of the dartboard as given in the question = 18 inches
Then
Radius of the dartboard will be = 18/2 inches
      = 9 inches
We already know that
Circumference of a circle = <span>2π</span>r
We also know that the value of <span>π = 3.14159
Then
Circumference of the dartboard = 2 * 3.14159 * 9 inches
      = 56.548 inches
      = 56.55 inches
So the radius of the dartboard is 9 inches and the circumference of the dartboard is 56.55 inches. I hope the procedure is absolutely clear to you.
</span>

3 0

3 years ago

Does anybody know how to find the volume of. a prism. please help

olganol [36]

V = [1/2 x 5 x side x apothem] x height of the prism.
V=[ 1/2 x 5 x 7.1 x 5] x 15
V= 88.75 x 15
V= 1331.25

7 0

3 years ago