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3 years ago

how long is a string reaching from the top of a 15 foot pole to a point 13 feet from the base of the pole?

Mathematics

2 answers:

umka21 [38]3 years ago

8 0

Pyth theorem
X^2=15^2+13^2
X=19.8

natima [27]3 years ago

6 0

The answer is 195 because you have to multiply

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What is the standard form of this expression? (4y-9)(2y-3)

Setler [38]

Answer:

-22y+27

Step-by-step explanation:

(4y-9)(2y-3)

<u>Distribute:</u>

8y-12y-18y+27

<u>Combine like terms:</u>

8y-12y-18y=-22y

-22y+27

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B40%7D" id="TexFormula1" title="\frac{28}{40}" alt="\frac{28}{40}" align="abs

Vitek1552 [10]

Answer:

7/10

Step-by-step explanation:

28/40=

14/20 =

7/10

8 0

3 years ago

Read 2 more answers

The ratio of the side length of a square to the square's perimeter is always 1 to 4.

makvit [3.9K]

Answer:

Side length of square (inches): 1 ,2, 4, 7

Perimeter of square (inches). : 4, 8, 16, 28

Step-by-step explanation:

Ratio is 1:4 (S:P).

So:

#:8, #=2. This is because 2/8 and 1/4 both = 0.25.

#:16, #=4. 4/16=0.25

7:#, #=28. 4x7=28, 7/28=0.25

Hope that helps

7 0

2 years ago

-(-8)= can someone please solve this? Is a fill in the blank.

butalik [34]

Answer:

Step-by-step explanation:

when multiplying two negative, it turn into a positive.

7 0

3 years ago

Read 2 more answers

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

4 years ago