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2 years ago

can someone explain why it keeps saying my comments violate guidelines when i don’t say anything that shouldn’t be said?

Mathematics

2 answers:

sineoko [7]2 years ago

8 0

it could be that someone reported u account.

Likurg_2 [28]2 years ago

4 0

Answer:

Step-by-step explanation:

i think it’s just a glitch i keep getting those messages because i answered a question that got deleted

You might be interested in

How many solutions are there to the equation below? |x| = -31

ella [17]

Break down the problem into these two equations;

x = -31

-x = -31

Solve for the 1st equation; x = -31

x = -31

Solve for the 2nd equation; -x = -31

x = 31

Collect all solutions

x = ±31

Check the solution

When x = -31, the original equation; |x| = -31 does not hold true. Thus, we will drop x = -31 from the solution set.

Check the solution again;

When x = 31, the original equation |x| = -31 does not hold true as well. Thus we will drop x = 31 from the solution set.

Therefore,

No solution exists to this equation.

5 0

3 years ago

Read 2 more answers

For what values of x is XP 36-5x true?

KiRa [710]

I am not sure this is the answer but I believe it’s 4 and -9

4 0

2 years ago

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,

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