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bazaltina [42]
3 years ago
8

A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it is replaced and another ball is dr

awn. This process goes on indefinitely. What is the probability that of the first 4 balls drawn, exactly 2 are white?
Physics
1 answer:
Ahat [919]3 years ago
5 0

Answer:0.375

Explanation:

Given

An Urn contains 3 White and 3 black balls

Ball is replaced after it is drawn

Using Binomial Distribution as trials are finite

n=4 i.e. 4 balls are drawn

Probability of getting white ball (p)=\frac{1}{2}

Probability of getting a Non-white ball(q)=\frac{1}{2}

P(X=r)= ^nC_r(p)^r(q)^{n-r}

For Exactly 2 white balls

P(X=2)=^4C_2(\frac{1}{2})^{2}(\frac{1}{2})^{2}

P(X=2)=\frac{4!}{2!\cdot 2!}\times \frac{1}{2^4}

P(X=2)=\frac{3}{8}

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3 years ago
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
4 years ago
Need help solving this
Nostrana [21]

Answer:

cant see picture

Explanation:

5 0
3 years ago
An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and
Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

\Delta P.E=mg(h_{2}-h_{1})

\Delta P.E=10000\times9.8\times(10000-0)

\Delta P.E=10000\times9.8\times10000

\Delta P.E=980000000\ J

\Delta P.E=980\ MJ

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)

\Delta K.E=148298642\ J

\Delta K.E=148.3\ MJ

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0
3 years ago
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