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3 years ago

A pair of dice are rolled. What is the probability of rolling 4 or greater?

1 answer:

6 0

Answer: option D: 11/12

Explanation:

1) The sample space is formed by 36 pairs: 6×6 = 36.

2) Roolling 4 or more is the same that not rolling 3 or less

3) Rolling 3 or less are (1,1), (1,2), and (2,1). Those are 3 outcomes out of 36.

4) Therefore, not rolling 3 or less (the same that rolling 4 or more) has 36 - 3 = 33 outcomes out of 36.

5) The probability of 33 out of 36 is 33/36

6) Simplify the fraction 33/36 = 11/12

And that is the answer.

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3 years ago

A circle has a radius of 4. An arc in this circle has a central angle of 288°.

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3 years ago

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Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region $\mathcal T$ by computing the double integral of the density function over $\mathcal T$ :

$\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy$

The boundary of $\mathcal T$ is determined by a set of lines in the $x,y$ plane. One way to describe the region $\mathcal T$ is by the set of points,

$\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}$

So the mass is

$\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx$

$=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52$

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3 years ago