Two six-sided dice are rolled. Find the probability that the sum of the two faces that turn up is between 1 and 14.

Mathematics

1 answer:

lina2011 [118]3 years ago

4 0

Answer:

100%

Step-by-step explanation:

100 percent.

No matter what, the sum of the two faces that turn up will be between 1 and 14.

This means that it will occur, 100 percent of the time, or 10 out of 10 times.

Since each die is 1-6, the sum of the two faces will always be between 2 and 12.

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Hunter-Best [27]

Let's simplify the following expression:

$\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}$

We get,

$\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}$ $\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}$