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grigory [225]
3 years ago
9

Twenty dash three people purchase raffle tickets. Three winning tickets are selected at random. If first prize is ​$1000​, secon

d prize is ​$500​, and third prize is ​$100​, in how many different ways can the prizes be​ awarded?
Mathematics
1 answer:
Y_Kistochka [10]3 years ago
5 0

Answer: 10626

Step-by-step explanation:

Given : The total number of people purchase raffle tickets=23

The total number of prizes = 3

Since , order matters here , so we use Permutations.

The permutation of n things taken r at a time is given by :-

^nP_r=\dfrac{n!}{(n-r)!}

Then, the number of ways the prizes can be distributed :-

^{23}P_{3}=\dfrac{23!}{(23-3)!}\\\\=\dfrac{23\times22\times21\times20!}{20!}=10626

Hence, the prizes can be distributed in 10626 ways.

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Am I dealing with harmonic stuff (trig)?
pochemuha

You just need to solve for when y=0:

\dfrac{\cos8t-9\sin8t}4=0\implies\cos8t-9\sin8t=0

\implies\cos8t=9\sin8t

\implies\dfrac19=\dfrac{\sin8t}{\cos8t}=\tan8t

\implies8t=\tan^{-1}\dfrac19+n\pi

\implies t=\dfrac18\tan^{-1}\dfrac19+\dfrac{n\pi}8

where n is any integer. We only care about when 0\le t\le1, which happens for n\in\{0,1,2\}.

t=\dfrac18\tan^{-1}\dfrac19\approx0.01

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4 years ago
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6 0
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Step-by-step explanation:

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3 years ago
Find the solutions of the quadratic equation 14x^2+9x+10=014x
Vesnalui [34]

Answer:

Option B. x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

14x^{2}+9x+10=0

so

a=14\\b=9\\c=10

substitute in the formula

x=\frac{-9(+/-)\sqrt{9^{2}-4(14)(10)}} {2(14)}

x=\frac{-9(+/-)\sqrt{-479}} {28}

Remember that

i=\sqrt{-1}

substitute

x=\frac{-9(+/-)\sqrt{479}i} {28}  

x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

5 0
3 years ago
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