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Strike441 [17]
3 years ago
10

Ill mark brainlest if correct

Mathematics
1 answer:
ehidna [41]3 years ago
4 0
Same here , like somebody please answer thanks
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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
kompoz [17]

Answer:

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

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

and \sigma_{{\overline}{Y}} = 0.38

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 .

Let,

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}

                            = 3.5       ∀ i = 1(1)10 and j = 1(1)20

and,

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

                                = \frac {1 + 4 + 9 + 16 + 25 + 36}{6}

                                = \frac {91}{6}

                                \simeq 15.166667

so, Var(X_{ij} = (E(X^{2}_{ij} - {(E(X_{ij})}^{2})

                                    \simeq 15.166667 - 3.5^{2}

                                    = 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

and,

       \sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}}                                             = \frac {\sigma_{X_{ij}}}{\sqrt {20}}                                             = \frac {1.7078261036}{\sqrt {20}}                                            [tex]\simeq 0.38

Hence, the option which best describes the distribution being simulated is given by,

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

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

and \sigma_{{\overline}{Y}} = 0.38

                                   

6 0
3 years ago
Help plz..And No links!! I repeat No links!!
kompoz [17]

Answer:

6:09

Step-by-step explanation:

clocks are cool and fun

5 0
3 years ago
Read 2 more answers
Can I please have some HELP!!! (Brainllest)
Solnce55 [7]

Answer:

its the last one I think.

3 0
3 years ago
Solve the following system of equations by linear combination: 5x 2y = 14 y = x – 7
natta225 [31]
I would use subsitution
y=x-7
sub x-7 for every x
5x+2(x-7)=14
5x+2x-14=14
add 14 to both sides
7x=28
divide both sides by 7
x=4
sub
y=x-7
y=4-7
y=-3

x=4
y=-3
(4,-3)
4 0
3 years ago
If someone can help me with this question please?
Katyanochek1 [597]

9514 1404 393

Explanation:

The three Reasons tell you what to look for to put in the Statement blank.

1. We are given that RE = 2AR and RT = 2GR.

2. The only vertical angles in the figure are ...

  ∠GRA ≅ ∠TRE

3. Using the given relation between the sides, we can write the proportion ...

  RE/RA = RT/RG = 2

It is nice, though maybe not absolutely essential, to write the segment names in order of corresponding vertices.

4. Having shown that two sides are proportional and the angle between them is congruent, we can claim similarity using the SAS Theorem.

8 0
3 years ago
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