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1 year ago

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You and a friend are playing a game of chance. Every time you roll a 1 or 2 you are successful, and your friend will pay you $1.

Every time you roll (using a fair die) a 3, 4, 5 or 6, you must pay your friend $2. If 71% of your rolls are successful and 29% of your rolls are unsuccessful, how much money do you expect to have earned/owed by the end of the game

1 answer:

Viktor [21]1 year ago

8 0

Considering the mean of a discrete distribution, you are expected to earn $0.13 by the end of the game.

<h3>What is the mean of a discrete distribution?</h3>

The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.

Since 71% of your rolls are successful and 29% of your rolls are unsuccessful, the distribution of your earnings is:

P(X = 1) = 0.71.

P(X = -2) = 0.29

Hence the expected value is:

E(X) = 1 x 0.71 - 2 x 0.29 = 0.71 - 0.58 = 0.13

More can be learned about the mean of a discrete distribution at brainly.com/question/24245882

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Find a polynomial with integer coefficients that satisfies the given conditions. R has degree 4 and zeros 3 − 3i and 2, with 2 a

dolphi86 [110]

Answer:

The required polynomial is $P(x)=x^4-10x^3+46x^2-96x+72$ .

Step-by-step explanation:

If a polynomial has degree n and $c_1,c_2,...,c_n$ are zeroes of the polynomial, then the polynomial is defined as

$P(x)=a(x-c_1)(x-c_2)...(x-x_n)$

It is given that the polynomial R has degree 4 and zeros 3 − 3i and 2. The multiplicity of zero 2 is 2.

According to complex conjugate theorem, if a+ib is zero of a polynomial, then its conjugate a-ib is also a zero of that polynomial.

Since 3-3i is zero, therefore 3+3i is also a zero.

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$R(x)=((x-3)+3i)((x-3)-3i)(x-2)^2$

$R(x)=(x-3)^2-(3i)^2((x-3)-3i)(x-2)^2$ $[a^2-b^2=(a-b)(a+b)]$

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Therefore the required polynomial is $P(x)=x^4-10x^3+46x^2-96x+72$ .

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2 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

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$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

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$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

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$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

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Distribute:

$\sqrt{3}-2$

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2 years ago