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SOVA2 [1]
3 years ago
8

What is the probability you roll doubles on two 6 sides dice please show work

Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

1/6

Step-by-step explanation:

We can start by determining all the different combinations we can get.

We are using 6 sided dices, so we have a chance of getting 6 different numbers for each dice.

If we only have 1 dice, the our possible values would be:

1, 2, 3, 4, 5, 6

However, we have 2 dices. As such, our combinations have been multiplied, and now our possibilities are:

(1 , 1) , (1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (1 , 6) , (2 , 1) , (2 , 2) , (2 , 3) , (2 , 4) , (2 , 5) , (2 ,6), and so on for 3, 4, 5, and 6

We can find the number of combinations by simply multiplying the amount of possible values for each dice.

Since both dices have 6 possible values, we multiply 6 by 6, and the amount of possible combinations is 36.

Now, let's determine the possibility for getting a double.

First, let's determine how many doubles we can get:

(1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) , (5 , 5) , (6 ,6)

There are 6 different combinations for rolling a double.

So, our probability of rolling a double on two 6 sided dices is 6 out of 36

We can simplify that to 1 out of 6.

So the probability of rolling a double on two 6 sided dices is 1/6

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Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4
Kay [80]

Answer:

<h2>c. x = -4 or x = 9</h2>

Step-by-step explanation:

\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}

Domain:

x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3

solution:

\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}

use <em>(a - b)(a + b) = a² - b²</em>

\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}

multiply both sides by (x - 3) ≠ 0

\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}

cancel (x - 3)

\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3

subtract \frac{4(x-3)}{x} from both sides

\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}

cross multiply

(4)(x)=(x+3)(-x+12)

use FOIL

4x=(x)(-x)+(x)(12)+(3)(-x)+(3)(12)\\\\4x=-x^2+12x-3x+36

subtract 4x from both sides

0=-x^2+12x-3x+36-4x

combine like terms

0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36

change the signs

x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0

The product is 0 if one of the factors is 0. Therefore:

x-9=0\ \vee\ x+4=0

x-9=0            <em>add 9 to both sides</em>

x=9\in D

x+4=0          <em>subtract 4 from both sides</em>

x=-4\in D

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