Help please!!!!!! asap

The probability is defined as the chances of happening of any random event. or the possibility of happening of any event is called as the probability.

It is given in the question that there are two colours in the spinner and the dice have 6 faces from {1 to 6} numbers on it.

The sample outcomes with the spinner and the dice will be as follows:-

S= {(R,1),(R,2)(R,3)(R,4(R,5)(R,6),(B,1)(B,2)(B,3)(B,4)(B,5)(B,6).

Hence answer will be {(R,1),(R,2)(R,3)(R,4(R,5)(R,6),(B,1)(B,2)(B,3)(B,4)(B,5)(B,6).

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5 0

2 years ago

Find the volume of the solid formed by rotating the region bounded by the given curves about the indicated axis. Y = 1x, y = 1,

Goryan [66]

The disk method will only involve a single integral. I've attached a sketch of the bounded region (in red) and one such disk made by revolving it around the y-axis.

Such a disk has radius x = 1/y and height/thickness ∆y, so that the volume of one such disk is

π (radius) (height) = π (1/y)² ∆y = π/y² ∆y

and the volume of a stack of n such disks is

$\displaystyle V_n = \sum_{i=1}^n \pi {y_i}^2 \Delta y$

where $y_i$ is a point sampled from the interval [1, 5].

As we refine the solid by adding increasingly more, increasingly thinner disks, so that ∆y converges to 0, the sum converges to a definite integral that gives the exact volume V,

$\displaystyle V = \lim_{n\to\infty} V_n = \int_1^5 \frac{\pi}{y^2} \, dy$

$V = -\dfrac\pi y\bigg|_{y=1}^{y=5} = \boxed{\dfrac{4\pi}5}$

8 0

2 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago