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3 years ago

8

You are challenged to a lucky draw game. If you draw a face card (K, Q, J) from a standard deck of cards, you earn 10 points. If

you draw any other card, you lose 2 points. What is the expected value of a draw?

2 answers:

sveticcg [70]3 years ago

7 0

There are 3 face cards for each of the 4 suits, leading to a total possible gain of 120, there are 40 other cards leading to a total possible loss of 80.
120 - 80 = 40.
40 / 52 (total number of cards) = about .77

Final Answer:
The expected value of a draw is positive 0.77 points.
Hope I helped :)

Natali [406]3 years ago

6 0

Answer:0.77

Step-by-step explanation:

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The sum of four times a number and five times another number is -28. The sum of nine times the first number and two times the se

dedylja [7]

Answer:

-8

Step-by-step explanation:

If we let x and y represent the first and second numbers, respectively, we can write the problem statement as the equations ...

4x +5y = -28
9x +2y = 11

We can multiply the first equation by -9 and add 4 times the second equation to get an equation in y alone:

-9(4x +5y) +4(9x +2y) = -9(-28) +4(11)

-37y = 296 . . . . . simplify

296/-37 = y = -8 . . . . . divide by the coefficient of y

The second number is -8.

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3 years ago

If the diagonal of a square is approximately 12.73, what is the length of its sides?

Alja [10]

Answer:

The length of the sides of the square is 9.0015

Step-by-step explanation:

Given

The diagonal of a square = 12.73

Required

The length of its side

Let the length and the diagonal of the square be represented by L and D, respectively.

So that

D = 12.73

The relationship between the diagonal and the length of a square is given by the Pythagoras theorem as follows:

$D^{2} = L^{2} + L^{2}$

Solving further, we have

$D^{2} = 2L^{2}$

Divide both sides by 2

$\frac{D^{2}}{2} = \frac{2L^{2}}{2}$

$\frac{D^{2}}{2} = L^2$

Take Square root of both sides

$\sqrt{\frac{D^{2}}{2}} = \sqrt{L^2}$

$\sqrt{\frac{D^{2}}{2}} = L$

Reorder

$L = \sqrt{\frac{D^{2}}{2}}$

Now, the value of L can be calculated by substituting 12.73 for D

$L = \sqrt{\frac{12.73^{2}}{2}}$

$L = \sqrt{\frac{162.0529}{2}}$

$L = \sqrt{{81.02645}$

$L = 9.001469325$

$L = 9.0015$ (Approximated)

Hence, the length of the sides of the square is approximately 9.0015

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3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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3 years ago

Which expression is 8y - 5x + 13y equivalent to after being simplified?

Thepotemich [5.8K]

You combine like terms like 8y+13y which gives you 21y and don't forget the -5x so your simplified equation would end up being A. 21 - 5x

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Step-by-step explanation:

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