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shusha [124]
3 years ago
14

You are given 5 to 4 odds against tossing three heads heads with three​ coins, meaning you win ​$5 if you succeed and you lose ​

$4 if you fail. Find the expected value​ (to you) of the game. Would you expect to win or lose money in 1​ game? In 100​ games? Explain. Find the expected value​ (to you) for the game.- ​$___ ​(Type an integer or a decimal rounded to the nearest hundredth as​ needed.)
Mathematics
1 answer:
cestrela7 [59]3 years ago
8 0

Answer:

$-2.875 in 1 game

$-287.5 in 100 games

Step-by-step explanation:

As the probability of getting head in tossing a fair coin is 0.5, the probability of getting 3 heads with 3 coins is 0.5*0.5*0.5 = 0.125

So you have a 0.125 chance of winning and 0.875 chance of losing a game

The expected value for 1 game would be

0.125*5 - 0.875*4 = -2.875

On average you would be losing 2.875  per game. In 100 games you should expect to lose 287.5 dollar

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3 years ago
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Factor completely x^3 + 1/8
maksim [4K]

Answer:

((2 x + 1) (4 x^2 - 2 x + 1))/8

Step-by-step explanation:

Factor the following:

x^3 + 1/8

Put each term in x^3 + 1/8 over the common denominator 8: x^3 + 1/8 = (8 x^3)/8 + 1/8:

(8 x^3)/8 + 1/8

(8 x^3)/8 + 1/8 = (8 x^3 + 1)/8:

(8 x^3 + 1)/8

8 x^3 + 1 = (2 x)^3 + 1^3:

((2 x)^3 + 1^3)/8

Factor the sum of two cubes. (2 x)^3 + 1^3 = (2 x + 1) ((2 x)^2 - 2 x + 1^2):

((2 x + 1) ((2 x)^2 - 2 x + 1^2))/8

1^2 = 1:

((2 x + 1) ((2 x)^2 - 2 x + 1))/8

Multiply each exponent in 2 x by 2:

((2 x + 1) (2^2 x^2 - 2 x + 1))/8

2^2 = 4:

Answer:  ((2 x + 1) (4 x^2 - 2 x + 1))/8

5 0
3 years ago
Read 2 more answers
What is 4876510×7928670
myrzilka [38]
4876510*7928670  \text{Multiplication} 

\text{Solution = }38664238541700

5 0
3 years ago
I need help please and thank you
lys-0071 [83]

Answer: it’s 135

Step-by-step explanation:

4 0
3 years ago
PLSS HELP ILL GIVE YOU BRAINLIEST!!
uysha [10]

Answer:

Step-by-step explanation:

Q1)

Use Phythogoras theorem:

(AC)^2=(BC)^2+(AB)^2\\BC^2=AC^2-AB^2\\BC^2=6^2-4^2\\BC^2=36-16\\BC^2=20\\\\Apply\ square\ on\ both\ sides\\\\\sqrt{BC^2}=\sqrt{20}  \\BC=\sqrt{20} \\BC=\sqrt{10(2)} \\BC=\sqrt{5(2)(2)} \\BC=2\sqrt{5}

Q2)

Apply phythogoras theorem:

CD^2=CE^2+DE^2\\CD^2=7^2+9^2\\CD^2=49+81\\CD^2=130\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{CD^2} = \sqrt{130} \\\\CD=\sqrt{130} \\\\CD=11.4

Q3)

Apply phythogoras theorem again:

BC^2=BD^2+CD^2\\BC^2=7^2+13^2\\BC^2=49+169\\BC^2=218\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{BC^2} =\sqrt{218}  \\\\BC=\sqrt{218} \\\\BC=14.76

I have an attached an image for Question 2 for better understanding the length of DE in question equals 15 - 6 = 9

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3 years ago
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