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olga55 [171]
3 years ago
15

-41.3 x 10 x 1.7 x 10-3x 10Q2[2]Ed9.​

Mathematics
1 answer:
Len [333]3 years ago
3 0

Answer:

221

Step-by-step explanation:

use bidmas (brackets,indices, division,multiplication,addition,substraction)

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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

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
Where are the x-intercepts for f(x) = −4cos(x − pi over 2) from x = 0 to x = 2π?
yKpoI14uk [10]
Recall that to get the x-intercepts, we set the f(x) = y = 0, thus

\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2}  \right)\implies 0=cos\left(x-\frac{\pi }{2}  \right)
\\\\\\
cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2}  \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2}
\\\\\\
x-\cfrac{\pi }{2}=
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}

\bf -------------------------------\\\\
x-\cfrac{\pi }{2}=\cfrac{\pi }{2}\implies x=\cfrac{\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{2\pi }{2}\implies \boxed{x=\pi }\\\\
-------------------------------\\\\
x-\cfrac{\pi }{2}=\cfrac{3\pi }{2}\implies x=\cfrac{3\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{4\pi }{2}\implies \boxed{x=2\pi }
3 0
3 years ago
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