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AysviL [449]
3 years ago
9

In a certain card game you draw one card off a standard deck of 52 cards. If you draw a spade you get paid $12, if you draw a re

d Ace you get paid $20, and if you draw a red Queen you get paid $38. If you draw anything else, you get paid nothing. What should this game cost if it is to be a fair game? Use fractions in your work and then calculate the answer as a decimal rounded to 4 decimal places.
Mathematics
1 answer:
Mariulka [41]3 years ago
4 0

Step-by-step explanation:

In a standard deck of 52 cards, there are 2 red aces, 2 red Queens, and 13 spades.  That leaves 35 cards for everything else.

For the game to be fair, the cost must equal the expected value.  The expected value is the sum of each outcome times its probability.

C = (12) (13/52) + (20) (2/52) + (38) (2/52) + (0) (35/52)

C = 68/13

C ≈ 5.2308

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"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx

For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)

So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
7 0
3 years ago
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Answer:

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Step-by-step explanation:

You have to contribute and combine like terms and that is how you get your answer

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SOMEONE HELP ME PLEASE
OlgaM077 [116]

Step-by-step explanation:

If Angle 1 = 94°

Then

Angle 2 = 180°-94° = 86°

( Linear pairs or supplementary angles with angle 1 )

Angle 3 = 94°

( Corresponding angle with angle 1 )

Angle 4 = 86°

( Corresponding angle with angle 2 )

Angle 5 = 94°

( Exterior alternate angle with angle 1 or vertically opposite angle with angle 3 )

Angle 6 = 86°

( Alternate Interior angle with angle 2 or vertically opposite angles with angle 4 )

Angle 7 = 94°

( Alternate Interior angle with angle 3 and vertically opposite angles with angle 1)

Angle 8 = 86°

( Alternate exterior angle with angle 4 or vertically opposite angles with angle 2 )

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