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3 years ago

a fair number cube labeld 1,2,3,4,5,6 is rolled a hundred times. about how many times could you expect to roll a three?

1 answer:

3 0

So probability is (desired outcomes)/(total possitlble outcomes)
desired is 1 since 3 appaears once on the cube
6 total possible

so probaity of rolling a3 is 1/6

100 times
100 times 1/6=100/6=16 and 2/3 times
round up
17 times

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3 years ago

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2 years ago

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miss Akunina [59]

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$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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1st boat:

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$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

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$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

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$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

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$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

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$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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3 years ago

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pshichka [43]

The answer is 11
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Add the 2 from 2 1/3 to the nine
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2 years ago

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