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wariber [46]
3 years ago
14

A circle has an area of 201.0619. What is the diameter? (round to the nearest whole number)

Mathematics
2 answers:
bija089 [108]3 years ago
8 0

The diameter of the circle is approximately 16

jonny [76]3 years ago
4 0

Answer:

16

Step-by-step explanation:

d≈16

A Area

201.0619

Using the formulas

A=πr2

d=2r

Solving ford

d=2A

π=2·201.06

π≈16

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Step-by-step explanation:

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A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we
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Answer:

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

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Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

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X_{ij} = The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

E(X_{ij}) = \frac {1 + 2 + 3 + 4 + 5 + 6}{6}

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E(X^{2}_{ij} = \frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}

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                                \simeq 15.166667

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                                    = 2.91667

   and \sigma_{X_{ij}} = \sqrt {2.91667}[/tex                                            [tex]\simeq 1.7078261036

Now we get that,

 Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}

We get that Y_{j}'s are iid RV's ∀ j = 1(1)20

Let, {\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}

      So, we get that E({\overline}{Y}) = E(Y_{j})

                                                                 = E(X_{ij}  for any i = 1(1)10

                                                                 = 3.5

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C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

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