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3 years ago

11

What’s the anwser Simplify √3 •√21

1 answer:

Elis [28]3 years ago

4 0

Answer:

$3\sqrt{7}$

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Determine whether the lines are parallel, perpendicular, or neither. <br> x = 3<br> and y = 4

Katyanochek1 [597]

I think they’re perpendicular lines

5 0

3 years ago

An Olympic-sized swimming pool can hold approximately 6.60 x 105 gallons of water. Calculate the volume of the water in cubic me

shutvik [7]

The volume of the pool is 1.74*10^8 cubic meters

To solve this problem, we convert the volume of the water from gallons to m^3.

<h3 /><h3>Conversion of Unit</h3>

This is done to convert one unit of a substance to another unit of that particular substance. In this question, we are converting from gallon - meter cube

Data;

v = $6.60*10^5$

$1 gallon = 264.17m^3$

let's solve for v

$1gallon = 264.17m^3\\
6.60*10^5 gallon = x$

cross multiply both sides and solve for the volume in cubic meters

$x = 6.60*10^5 * 264.17\\
x = 174352200\\
x = 174.352*10^6m^3 = 1.74*10^8m^3$

The volume of the pool is 1.74*10^8 cubic meters

Learn more on conversion of unit here;

brainly.com/question/8426032

3 0

3 years ago

Semenov [28]

Answer:

$- {x}^{2} + x + 24$

Step-by-step explanation:

Subtract the functions:

$f - g = (36 - {x}^{2} ) - (8 - x)$

Simplify:

$(36 - {x}^{2} ) - (8 - x) \\ 36 - {x}^{2} - 8 + x \\ - {x}^{2} + x + 24$

7 0

4 years ago

Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

a. Then, for $\hat \theta_3$ to be an unbiased estimator ; Then:

$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

Thus; in order to minimize the variance of $\hat \theta_3$ ; then constant a can be determined as :

$V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2$

Using differentiation:

$\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0$

⇒

$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

$\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}$

This implies that

$\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0$

So, $V( \hat \theta_3)$ is minimum when $\hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}$

As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

4 0

4 years ago

WILL GIVE BRAINLIST SO PLZ HELP

DochEvi [55]

Hello!

D Would be your answer! Because At Midnight, the temperature was -8*f. At noon, 23*F, so

|-8-23| |-3||= 31 There you go! May I have Brainliest?

4 0

4 years ago