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nikdorinn [45]
3 years ago
13

In a class of 888, there are 333 students who have done their homework. If the teacher chooses 222 students, what is the probabi

lity that both of them have done their homework?
Mathematics
2 answers:
vlabodo [156]3 years ago
8 0
555 of the students haven’t did there work
TEA [102]3 years ago
7 0

Answer:

3/28

Step-by-step explanation:

3/8 * 2/7 = 6/56 = 3/28

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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
In ΔCDE, the measure of ∠E=90°, DC = 5, CE = 4, and ED = 3. What ratio represents the sine of ∠C?
lianna [129]

Answer:

As E is right angle , CD is hypotenuse , and ED is perpendicular and CE is the base .

Thus sinC = Perpendicular/ Hypotenuse

sinC = ED/CD

sinC = 3/5

3 1
3 years ago
What is the answer to 2X +1=-15
Over [174]
<span>2X +1=-15
Subtract 1 from both sides
2x=-16
Divide 2 on both sides
Final Answer: x=-8</span>
5 0
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What is 12 5​ as a mixed number in simplest form?<br> 25
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Answer:

5−4+4=16+4

Step-by-step explanation:

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inn [45]

Answer:

A. The initial value is 100; it represents money left over from last year's fundraiser

Step-by-step explanation:

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3 years ago
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