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Radda [10]
3 years ago
9

A die is rolled 20 times. given that three of the rolls came up 1, five came up 2, four came up 3, two came up 4, three came up

5, and three came up 6, how many different arrangements of the outcomes are there
Mathematics
1 answer:
Serjik [45]3 years ago
8 0

Solution:

Number of times a die is rolled = 20

1 - 3=A

2 - 5=B

3 - 4=C

4 - 2=D

5 - 3=E

6 -  3=F

Total number of arrangements of outcomes , when a dice is rolled 20 times given that 1 appear 3 times, 2 appears 5 times, 3 appear 4 times, 4 appear 2 times , 5 appear three times, and 6 appear 3 times

            = Arrangement of 6 numbers (A,B,C,D,E,F) in 6! ways and then arranging outcomes

= 6! × [ 3! × 5! × 4!×2!×3!×3!]

= 720 × 6×120×24×72→→[Keep in Mind →n!= n (n-1)(n-2)(n-3)........1]

= 895795200  Ways

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Hello!

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Answer:

z= 3.63

z for  significance level = 0.05 is ± 1.645

Step-by-step explanation:

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We formulate our null and alternative hypotheses as

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z= p`- p/ √p(1-p)/n

Putting the values

z= 0.5- 0.42/ √0.42*0.58/500

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Answer:

  • B. BC = 18 so ∆ABC ~ ∆DEF by SAS

Step-by-step explanation:

\bf \cfrac{AB}{DE}  =  \cfrac{BC}{EF}

  • \bf \cfrac{15}{10}  =  \cfrac{BC}{12}
  • \bf  \cfrac{3}{2}  =  \cfrac{BC}{12}
  • \bf BC =  \cfrac{3}{2}  \times 12
  • \bf 3 \times 6
  • \bf BC = 18
  • \bf \angle B = \angle \: E

So, ∆ABC ~ ∆DEF by SAS.

<h3><u>------------------------------------</u></h3>
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