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Sindrei [870]
3 years ago
11

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?

Mathematics
1 answer:
kupik [55]3 years ago
3 0

Answer:

There are 5,586,853,480 different ways to select the jury.

Step-by-step explanation:

The order is not important.

For example, if we had sets of 2 elements

Tremaine and Tre'davious would be the same set as Tre'davious and Tremaine. So we use the combinations formula.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?

Here we have n = 40, x = 12.

So

C_{40,12} = \frac{40!}{12!(28)!} = 5,586,853,480

There are 5,586,853,480 different ways to select the jury.

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The number of compact discs N purchased each year, in millions, can be
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Answer:

126.5 years

Step-by-step explanation:

N(t) = 384(1.13)^t

2,000,000,000 = 384(1.13)^t

(1.13)^t = 2,000,000,000/384

log [(1.13)^t) = log (2,000,000,000/384)

t × log 1.13 = log (2,000,000,000/384)

t = [log (2,000,000,000/384)]/(log 1.13)

t = 126.5

8 0
1 year ago
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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{4 {x}^{3}  + 6 {y}^{2} }}}}}

Step-by-step explanation:

\sf{5 {y }^{2}  + 5 - 4 - 4 {y}^{2}  + 5 {y}^{2}  - 4 {x}^{3}  - 1}

Collect like terms

⇒\sf{ 4 {x}^{3}  + 5 {y}^{2}  - 4 {y}^{2}  + 5 {y}^{2}  + 5 - 4 - 1}

⇒\sf{4 {x}^{3}  +  {y}^{2}  + 5 {y}^{2}  + 5 - 4 - 1}

⇒\sf{4 {x}^{3}  + 6 {y}^{2}  + 5 - 4 - 1}

Calculate

⇒\sf{4 {x}^{3}  + 6 {y}^{2}  + 1 - 1}

⇒\sf{4 {x}^{3}  + 6 {y}^{2} }

Hope I helped!

Best regards! :D

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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