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2 years ago

If I roll two six-sides dice, what is the probability that the sum of the numbers that show up on the dice is 3 or less?

Mathematics

1 answer:

liq [111]2 years ago

4 0

Answer:

1/12

Step-by-step explanation:

first we have to find all the possibilities of getting a sum of 3 or less: 1+1 or 1+2 and we count the second combination 2 times because the numbers can be on either of the dices so we have a total of 3 possibilities. all the possible pairimg of dice are 6*6=36 because each dice has 6 sides and we can get either of them. so the probability would be the chance of getting a sum of 3 or less divided by all the diff combination which equals 3/36 or 1/12 which is roughly around 8.3%

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d1i1m1o1n [39]

Answer:

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Step-by-step explanation:

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This can be written as: $2x^2 + 3x - 9 = 0$

To solve a quadratic equation of the form $ax^2 + bx + c = 0$ we use the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 2; b = 3; c = - 9

Therefore, the roots of the equation are:

$x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)}$

$\implies x = \frac{-3 \pm \sqrt{81}}{4}$

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We get two values of 'x', viz.,

x = $\frac{-3 + 9}{4}$ and $\frac{- 3 - 9}{4}$

$\implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4}$

⇒ x = -3, 3/2

Since the factors of the quadratic equation is asked, we write it as:

(x + 3)(x - $\frac{3}{2}$ ) = 0

because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.

Multiply (x + 3) and (x - $\frac{3}{2}$ to see that this indeed is the given quadratic equation.

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