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3 years ago

What is a possible volume for this cylinder if the diameter is 8 cm? Explain your reasoning.

Mathematics

1 answer:

lesantik [10]3 years ago

5 0

Answer:

The formula for the volume of a cylinder is pi*r^2*h

r=radius= 1/2 diameter

h= the height

pi=3.14...

just get the height and plug it in

Step-by-step explanation:

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Gina looks at the architectural plan of a four-walled room in which the walls meet each other at right angles. The length of one

masha68 [24]

If the shape of the room is a square, then the square root of the sum of the squares of the the two adjacent sides will give the diagonal.

i.e.

$\sqrt{17^2+17^2} = \sqrt{2(17^2)} \\ \\ = \sqrt{2(289)} = \sqrt{578} =24.04$

Since the square root of the sum of the squares of the the two adjacent sides is 24.04 and not 18.79, therefore, the shape of the room is not a square.

7 0

2 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

PLEASE ANSWER ASAP! WILL GIVE BRAINLIEST+12 POINTS!

VikaD [51]

Answer:

a¹⁰

Step-by-step explanation:

$(x^{m} )^{n} =x^{mn}$

So here we have (a⁵)², meaning we need to multiply the exponents to get a¹⁰

7 0

2 years ago

Which of the following are true statements about David Hilbert

adoni [48]

Need to know what the options are

8 0

2 years ago

Joy makes 6.5 litres of soup, correct to the nearest 0.5 litre. She serves the soup in 280 ml portions, correct to the nearest 1

Levart [38]

Answer:

Joy definitely does have enough soup for 22 people.

Step-by-step explanation:

Upper Bound: 285ml turn it to 0.285 L 10÷2=5

280ml :

Lower Bound: 275ml turn it to 0.275 L

Upper Bound: 6.75 L 0.5÷2=0.25

6.5 Litters :

Lower Bound:6.25 L

6.75÷0.275=24.54 or 25

6.25÷0.285=21.92 or 22

7 0

2 years ago

Read 2 more answers