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3 years ago

how is it possible for the expected value of a game to be losing 65 cents of the only outcomes are win $2 or lose $1

1 answer:

4 0

Let x be the probability of winning and let y be the probability of losing. Then we can write the following equation:
2x - y = -0.65 ..............(1)
the sum of the probabilities of winning and losing must equal 1.
x + y = 1............(2)
Rearranging (2), gives:
y = 1 - x ............(3)
Plugging the expression for y from (3) into (1), we get:
2x - 1 + x = -0.65
which simplifies to:
3x = 0.35
Therefore x = 0.117 and y = 0.883.
The expected value of the game will be a loss of 65 cents if the probability of winning is 0.117 and the probability of losing is 0.883.

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Match the factored form

Brut [27]

(3x - 8)(3x + 8) is 3.) 9x^2 - 64
(x + 6)^2 is 4.) x^2 + 12x + 36
(3x - 2)(x + 5) is 1.) 3x^2 + 13x - 10
5x(x - 1)(x + 7) is 2.) 5x^3 + 30x^2 - 35x
(x - 8)(x + 7) is 5.) x^2 - 10x + 16

I hope this helps!

7 0

3 years ago

Write 14/√2 + √50 in the form b√2 where b is an integer.

kifflom [539]

Answer:

$12 \sqrt{2}$

Step-by-step explanation:

Taking the terms one at a time,

$\frac{14}{ \sqrt{2} } \\ \frac{14 * \sqrt{2} }{ \sqrt{2} * \sqrt{2} } \\ \frac{14 \sqrt{2} }{2} \\ 7 \sqrt{2}$

$\sqrt{50} \\ \sqrt{25 * 2} \\ \sqrt{25}* \sqrt{2} \\ 5 * \sqrt{2} \\ 5 \sqrt{2}$

Adding the terms, you have;

$7 \sqrt{2} + 5 \sqrt{2} =(7 + 5) \sqrt{2} = 12 \sqrt{2}$

Hope it helps!

I'm out!

3 0

3 years ago

Read 2 more answers

For year y, the population of massachusetts was approximately what percent of the population of vermont?

kifflom [539]

Answer: 50%

Step-by-step explanation: 50%

5 0

2 years ago

Factor by grouping<img src="https://tex.z-dn.net/?f=%284y%5E3%20%2B%2028y%5E2%29%20%2B%20%28y%20%2B%207%29" id="TexFormula1" tit

Inessa05 [86]

We have the following expression given:

$(4y^3+28y^2)+(y+7)$

We can start selecting as common factor 4y^2 and we got:

$4y^2(y+7)+(y+7)$

Now we can select y+7 as common factor and we got:

$(y+7)\left\lbrack 4y^2+1\right\rbrack$

So then our final answer would be:

$(y+7)(4y^2+1)$

3 0

1 year ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0

1 year ago