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Nookie1986 [14]

3 years ago

13

tamara finds the sum of two number cubes rolled at the same time. the chart below shows all possible sums frim thr 36 possible c

ombinations when rolling two number cubes. how many times should tamara expect the sum of the two subes to be equal to 4 if she rolls the two number cubes 252 times?

1 answer:

Alla [95]3 years ago

8 0

Im notr sure but if we take the chances that the 4 sum has to be either (1.3) (3.1) (2.2) (2.2) that means its chance is 4/36 for each roll which is also 1/9 ,, we can multiply 1/9 times the roll times ( 252 ) we get =28

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Answer:

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A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

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then, substitute these in equation [1] to solve for c;

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Simplify:

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i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

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Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

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therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

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