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lyudmila [28]
3 years ago
13

When playing roulette at the Venetian casino in Las Vegas, a gambler is trying to decide whether to bet $5 on the number 27 or t

o bet $5 that the outcome is any one of these five possibilities: 0, 00, 1, 2, 3. From Example 6, we know that the expected value of the $5 bet for a single number is -26. For the $5 bet that the outcome is 0, 00, 1, 2, or 3, there is a probability of 5>38 of making a net profit of $30 and a 33>38 probability of losing $5.
a. Find the expected value for the ​$5 bet that the outcome is 00, 0, or 1.
b. Which bet is​ better: a $5 bet on the number 25 or a $5 bet hat the outcome is any one of the numbers 00, 0, or 1? Why?
Mathematics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that:

To bet $5 that the outcome is any one of these five possibilities: 0, 00, 1, 2, 3.

Let Y represent the Amount of net profit

Then, Y= {-5, 30}

The probability distribution of Y is:

Y                        -5                        30

P(Y=y)                \dfrac{33}{38}                        \dfrac{5}{38}

a)  The expected value of X is given by:

E[Y] =\sum y P(Y=y)= 30*\dfrac{5}{38}-5*\dfrac{33}{38}

=-0.39

= -39 \ cents

b)

On a bet of $5 on the number 25 we are expected to loose 24 cents.

While on a $5 bet that the outcome is any one of the numbers 0,00, or 1 we are expected to loose 39 cents.

Hence,  $5 bet on the number 27 is better. Because the expected loss is less in this bet

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ryzh [129]

Answer:

$104

Step-by-step explanation:

First, convert 60% to a decimal.

60% = 0.6

Now, multiply the cost to buy the camera ($65) by the decimal.

$65 × 0.6 = $39

Now, add this amount to the amount it cost to buy the camera.

$65 + $39 = $104

The selling price for the camera will be $104.

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Step-by-step explanation:

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3 years ago
There are 52 cards in a deck, and 13 of them are hearts. Four cards are dealt, one at a time, off the top of a well-shuffled dec
irga5000 [103]

Answer:

10.97%

Step-by-step explanation:

There are 52 cards.

13 of them, are hearts.

Then

52 - 13 = 39 cards are not hearts.

4 cards are drawn, we want to find the percent chance that the fourth card is a heart card, but no before.

So the first card can't be a heart card.

because the deck is well-shuffled, all the cards have the same probability of being drawn.

Then the probability of not getting a heart card, is equal to the quotient between the number of non-heart cards (39) and the total number of cards (52), then the probability is:

p₁ = 39/52

The second card also can't be a heart card, the probability is calculated in the same way than above, but now there are 38 non-heart cards and a total of 51 cards (because one card was already drawn) then the probability here is:

p₂ = 38/51

For the third card the reasoning is similar to the two above cases, here the probability is:

p₃ = 37/50

The fourth card should be a hearts card, the probability is computed in the same way than above, as the quotient between the number of heart cards in the deck (13) and the total number of cards in the deck (now there are 49 cards)

then the probability is:

p₄ = 13/49

The joint probability (the probability of these 4 events happening together) is equal to the product between the individual probabilities:

P = p₁*p₂*p₃*p₄

P = (39/52)*(38/51)*(37/50)*(13/49) = 0.1097

The percent chance is the above number times 100%

Percent =  0.1097*100% = 10.97%

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3 years ago
The equation of the line of best fit of a scatter plot is y = ñ5x − 2. What is the slope of the equation?
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5, because  y = mx + b is the equation, with m being the slope and b being the y-intercept.
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4 0
3 years ago
Help, I don’t know what I’m doing<br> Trigonometry
spin [16.1K]

Answer:

\frac{\sqrt{3} }{6}

Step-by-step explanation:

using the 30- 60- 90 triangle for exact values , then

tan60° = \sqrt{3} , and

cot60° = \frac{1}{tan60} = \frac{1}{\sqrt{3} }

cos60° = \frac{1}{2}

cot60° cos60°

= \frac{1}{\sqrt{3} } × \frac{1}{2}

= \frac{1}{2\sqrt{3} } ← rationalise the denominator by multiplying by \frac{\sqrt{3} }{\sqrt{3} }

= \frac{1}{2\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} }

= \frac{\sqrt{3} }{6}

5 0
2 years ago
Read 2 more answers
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