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Fed [463]
3 years ago
10

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

ompletion. If the first queue has a time quantum of 5ms and each queue thereafter has a time quantum that is twice as large as the previous one, how many times will the job be interrupted, and on which queue will it finish its execution? Explain how much time it spends in each queue.
Physics
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

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A company randomly selects 100 light bulbs every day for 40 days from its production process. If 600 defective light bulbs are f
lutik1710 [3]

Answer:

3 sigma lower control limit = 0.0429

Explanation:

Given.

n = 100

days = 100

Number of defective bulbs = 600 defective bulbs

Let p = Process Average

p = 600/(100*40)

P = 600/4000

P = 0.15

q = 1 - p

q = 1 - 0.15

q = 0.85

3 sigma lower limit = p - 3*√(pq/n)

Using the above formula

Substitute in the values

3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)

3 sigma lower control limit= 0.15 - 3√0.001275

3 sigma lower control limit = 0.15 - 3* 0.035707142142714

3 sigma lower control limit = 0.15 - 0.107121426428142

3 sigma lower control limit = 0.04287857357185

3 sigma lower control limit = 0.0429 ---- approximated

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3 years ago
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What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.
Paha777 [63]

Answer:

v_o=2503.08\frac{m}{s}

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

v_o=\sqrt{\frac{GM}{r}}

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}

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3 years ago
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In a circuit a battery is used to
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The power of batteries in a circuit. The key functions of a battery and bulb in a circuit are explained. A battery is a source of energy which provides a push - a voltage - of energy to get the current flowing in a circuit. A bulb uses the electrical energy provided by the battery, but does not use current.

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3 years ago
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Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

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And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

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