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3 years ago

TAKE MY POINTS YOU PEOPLE"S

Mathematics

2 answers:

Artemon [7]3 years ago

5 0

Answer:

thanks? i am not understanding the point of this

Nina [5.8K]3 years ago

3 0

Tyyyyyyyyyyyyyyyyyyyy

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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

Which of the following p values will lead us to reject the null hypothesis if the level of significance equals .05?

V125BC [204]

Answer:

So then our significance level is $\alpha=0.05$ and we need to remember these two conditions:

1) If the p value $p_v$ we have enough evidence to reject the null hypothesis at the significance level given

2) If the p value $p_v \geq \alpha$ we have enough evidence to FAIL reject the null hypothesis at the significance level given

And baed on the options we see that the only possibility would be:

d. 0.015

Step-by-step explanation:

We want to know for which value we would REJECT the null hypothesis.

So then our significance level is $\alpha=0.05$ and we need to remember these two conditions:

1) If the p value $p_v$ we have enough evidence to reject the null hypothesis at the significance level given

2) If the p value $p_v \geq \alpha$ we have enough evidence to FAIL reject the null hypothesis at the significance level given

And baed on the options we see that the only possibility would be:

d. 0.015

7 0

3 years ago

How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

FromTheMoon [43]

This is essentially asking how many different ways to partition 6 into 3 segments.

I am assuming "no ball in a box" is not allowed.

6 can be partitioned as

(4,1,1), (3,2,1), and (2,2,2)

So, calculate each partition, we get

(6 choose 4) + (6 choose 3)*(3 choose 2) + (6 choose 2) * (4 choose 2)

= 15 + 20*3 + 15*6

=165

5 0

3 years ago

What is the volume of the figure?

FromTheMoon [43]

It’s going to be the area of the triangle on the side times the length, so 0.5 x 36 x 15 x 48 = 12960

7 0

3 years ago

6x + -2y² + 4xy² + 3x² + 5xy² simplify it

r-ruslan [8.4K]

$\large\displaystyle\text{$\begin{gathered}\sf 6x+-2y^{2}+4xy^{2}+3x^{2} +5xy^{2} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Group\:like\:terms \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf =3x^2+6x-2y^2+4xy^2+5xy^2 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Sum\:similar\:elements:\:4xy^2+5xy^2=9xy^2 \end{gathered}$}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf 3x^2+6x-2y^2+9xy^2 \end{gathered}$} }$

7 0

2 years ago

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