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3 years ago

A bag had 3 pink marbles and 4 purple marbles find the probability of selecting 3 pink marbles without replacement

1 answer:

Klio2033 [76]3 years ago

7 0

There are 7 marbles in the bag.
The probability of drawing a pink one is (3/7) .

Then you have 1 pink one, and there are 6 marbles in the bag.
The probability of drawing a pink one is (2/6) .

Then you have 2 pink ones, and there are 5 marbles in the bag.
The probability of drawing a pink one is (1/5) .

The probability of all 3 draws being successful is

(3/7) (2/6) (1/5) = 6/210 = 1/35 = about 2.86% (rounded)

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Find the moment of inertia about the y-axis of the thin semicirular region of constant density

arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

$\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA$

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

$\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_x = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about the y-axis will be

$\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about O will be

$\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4$

More about the rotational inertia link is given below.

brainly.com/question/22513079

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