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Ray Of Light [21]
3 years ago
10

Consider the game of independently throwing three fair six-sided dice. There are six combi- nations in which the three resulting

numbers can sum up to 9, and also six combinations in which they can sum up to 10: 9 = 1 + 2 + 6 =1+3+5 = 1+4+4 = 2+ 3+ 4 = 2+2 +5 = 3+3+3 10 = 1+3+ 6 = 1+ 4+ 5 = 2+2+6 = 2+3+5 = 2 + 4 + 4 = 3+3+4 This seems to suggest that the chances of throwing 9 and 10 should be equal. Yet, a certain gambler in Florence in the early XVII century (most likely, the Grand Duke of Tuscany Cosimo II de Medici) noticed that in practice it is more likely to throw 10 than 9. Show that the probability to get a total of 10 is, in fact, larger than the probability to get a total of 9. (This problem was solved for the duke by Galileo Galilei.)
Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0

Answer:

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

  1. 1+3+ 6,
  2. 1+ 4+ 5,
  3. 2+2+6,
  4. 2+3+5,
  5. 2 + 4 + 4,
  6. 3+3+4

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

  1. 1+3+ 6 : 6 permutations
  2. 1+ 4+ 5 : 6 permutations
  3. 2+2+6: 3 permutations
  4. 2+3+5: 6 permutations
  5. 2 + 4 + 4: 3 permutations
  6. 3+3+4: 3 permutations

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

  1. 1 + 2 + 6: 6 permutations
  2. 1+3+5: 6 permutations
  3. 1+4+4: 3 permutations
  4. 2+ 3+ 4: 6 permutations
  5. 2+2 +5: 3 permutations
  6. 3+3+3: 1 permutation.

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

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Solve for a<br> y = ax2 + bx + c
Sidana [21]

Answer:

Non-answer

Step-by-step explanation: Don't like my answer? Let me know!

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2"   was replaced by   "x^2".  

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    y-(a*x^2+b*x+c)=0  

STEP

1

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Equation at the end of step 1

 y - ax2 - xb - c  = 0  

STEP

2

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Solving a Single Variable Equation

2.1     Solve   y-ax2-xb-c  = 0

In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.

We shall not handle this type of equations at this time.

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3 years ago
Karen correctly answered 85% of the questions on her math test. If she misses 6 problems how many questions were on the test?
Brums [2.3K]
100 - 85 =  15%

15% = 6
1% = 0.4
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Answer: There were 40 questions.
8 0
3 years ago
How can the numbers 3 7 12 2 make 11​
Gala2k [10]

Answer:

Step-by-step explanation:

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3 years ago
Read 2 more answers
Porfabor necesito ayuda en la esta pregunta. ¿Encuentra cuatro pares ordenados de la siguiente función? f(x) = X3 – 2X2 – 2
zlopas [31]

Answer:

(0, -2), (1, -3), (2, -2) y (3, 7) son pares ordenados de f(x) = x^{3}-2\cdot x^{2}-2.

Step-by-step explanation:

Un par ordenado es un elemento de la forma (x,f(x)), donde x es un elemento del dominio de la función, mientras f(x) es la imagen de la función evaluada en x. Entonces, un par ordenado que está contenido en la citada función debe satisfacer la siguiente condición:

La imagen de la función existe para un elemento dado del dominio. Esto es:

x \rightarrow f(x)

Dado que f(x) es una función polinómica, existe una imagen para todo elemento x. Ahora, se eligen elementos arbitrarios del dominio para determinar sus imágenes respectivas:

x = 0

f(0) = 0^{3}-2\cdot (0)^{2}-2

f(0) = -2

(0, -2) es un par ordenado de f(x) = x^{3}-2\cdot x^{2}-2.

x = 1

f(1) = 1^{3}-2\cdot (1)^{2}-2

f(1) = -3

(1, -3) es un par ordenado de f(x) = x^{3}-2\cdot x^{2}-2.

x = 2

f(2) = 2^{3}-2\cdot (2)^{2}-2

f(2) = -2

(2, -2) es un par ordenado de f(x) = x^{3}-2\cdot x^{2}-2.

x = 3

f(3) = 3^{3}-2\cdot (3)^{2}-2

f(3) = 7

(3, 7) es un par ordenado de f(x) = x^{3}-2\cdot x^{2}-2.

(0, -2), (1, -3), (2, -2) y (3, 7) son pares ordenados de f(x) = x^{3}-2\cdot x^{2}-2.

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