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Hunter-Best [27]
3 years ago
14

Two state parks, Elm Grove and Cathedral Rock, each recorded the number of visitors to the park each day for 15 days. The parks

made line plots from the data recorded. What statement about the two plots’ distributions is true? The degree of overlap between the two distributions is low. The degree of overlap between the two distributions is high. There is no overlap between the two distributions. The degree of overlap between the two distributions is moderate.

Mathematics
2 answers:
Oksanka [162]3 years ago
6 0

Answer:

which one is correct because I am taking the test right know and I need the answer!

Step-by-step explanation:

please tell me the anTwo state parks, Elm Grove and Cathedral Rock, each recorded the number of visitors to the park each day for 15 days. The parks made line plots from the data recorded.

Which statements about the two plots' are true?

Choose each correct statement.

The range of visitors in a day to Elm Grove is greater than the range of visitors in a day to Cathedral Rock.

The greatest number of visitors in a day to Elm Grove is greater than the least number of visitors in a day to Cathedral Rock.

The range of visitors in a day to Cathedral Rock is equal to the range of visitors in a day to Elm Grove.

The least number of visitors in a day to Cathedral Rock is equal to the greatest number of visitors in a day to Elm Grove

Two line plots with the same scale and interval, from zero to six hundred, with tick marks every fifty units. The plots are stacked vertically for comparison. Values appear as x marks above the line. Each plot has fourteen data values. Upper plot is labeled Elm Grove. Upper plot data values are one x mark above fifty, two x marks above one hundred, three x marks above one hundred fifty, four x marks above two hundred, three x marks above two hundred fifty, one x mark above three hundred, and one x mark above three hundred fifty. Lower plot is labeled Cathedral Rock. Lower plot data values are one mark above three hundred, one makr above three hundred fifty, two marks above four hundred, three marks above four hundred fifty, four marks above five hundred, three marks above five hundred fifty, and one mark above six hundred.swer to this question:

ICE Princess25 [194]3 years ago
3 0

Answer:

The degree of overlap between the two distributions is moderate.

Step-by-step explanation:

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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}

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What is the relationship between x and y?
DedPeter [7]
Y=x is the answer alone, since it only crosses the origin
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3 years ago
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What are ALL the 100 digits of pi?
Sliva [168]

Answer:

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679

7 0
3 years ago
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When completely filled, a spherical storage tank
mamaluj [8]

Answer:

radius: 12

Given the following:

Volume of sphere: 7,240

Formula of volume of sphere:

Volume = 4/3 π (radius)³

Find radius:

  • 4/3 π (radius)³ = 7,240

<u>change sides</u>

  • (radius)³ = 7,240/(4π/3)

<u>simplify</u>

  • (radius)³ = 1728.422682

<u>cube root on both sides</u>

  • radius = ∛1728.422682

<u>simplify</u>

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2 years ago
(1×10,000)+(3×100)+(5×10)+(9× 10 1 ​ )+(8× 100 1 ​ ) as a decimal number.
andrey2020 [161]

Answer: 19,267.0


Step-by-step explanation:

1. You have the following information given in the problem above:

(1*10,000)+(3*100)+(5*10)+(9*101)+(8*1001)

2. To solve this exercise, first you must multiply the number that ar inside the parentheses and then you must add all of them, as you can see below:

10,000+300+50+909+8008=19,267

3. Rewriting is as a decimal, you have that the answer is:

19,267.0

8 0
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