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Ira Lisetskai [31]
2 years ago
4

Your friend plays a game of chance based on the sum of the rolls of three six-sided dice. Clearly the totals can range between t

hree (all ones) and eighteen (all sixes), however, your friend decides he wants to know the probability of each value appearing. Rather than do the mathematics he decides to find out by trials. He rolls the dice 1,000 times and counts the number of threes that occur. Then he rolls the dice another 1,000 times and counts the fours and so forth. You decide you’d rather him play Frisbee Golf with you and so you decide to model his process using a program. Writing and running the program will be faster than his method of finding the probabilities. Your program should output the count of each number in sequence, just as he would do.
Engineering
1 answer:
Alisiya [41]2 years ago
3 0

Answer:

import java.util.*;

class ProbabilityOfRollingThreeDice

{

   public static void main(String[] args)

   {

      Random rnd = new Random();

      System.out.print("Enter a random seed: ");

      long seed = new Scanner(System.in).nextLong();

      rnd.setSeed(seed);

      for(int i = 3; i <= 18; i++)

      {

         int count = 0;

         

         for(int j = 0; j < 1000; j++)

         {

            int roll1 = rnd.nextInt(6) + 1;

            int roll2 = rnd.nextInt(6) + 1;

            int roll3 = rnd.nextInt(6) + 1;

            int sum = roll1 + roll2 + roll3;

            if(sum == i)

                 count++;

         }      

         System.out.println("The value " + i + " occurred " + count + " times.");

      }

   }

}

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B is correct! in this senecio
5 0
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You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac
torisob [31]

Answer:

B, its the only valid answer

5 0
2 years ago
The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc
Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

\theta = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑f_{y} = 0, ∑M = 0

If ∑M_{A} = 0

⇒F_{BC}×0.9 - P × 0.6 = 0

⇒F_{BC}×3 - P × 2 = 0

⇒F_{BC} = \frac{2P}{3}

If ∑M_{B} = 0

⇒F_{AD}×0.9 = P × 0.3

⇒F_{AD} ×3 = P

⇒F_{AD} = \frac{P}{3}

Now,

Area, A = \frac{\pi }{4} X (0.012)^{2} = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , \delta = \frac{P l}{A E}

Now,

\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 9.1626 × 10⁻⁹ P

\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 1.83253 × 10⁻⁸ P

Given that,

\theta = 0.015°

⇒\theta = 2.618 × 10⁻⁴ rad

So,

\theta =  \frac{\delta_{BC} - \delta_{AD}}{0.9}

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0
2 years ago
Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige
Lera25 [3.4K]

Answer:

A) COP = \frac{16.97}{9.8} = 1.731

B) P_{IN} = 0.4763

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

T_1 = 28 degree celcius

T_2 = 12 degree celcius

\dot m = 1140 kg/h

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

A) Q = \dot m Cp \Delta T

        = 1140 \times 3.35\times (28-12) = 61,104 kJ/h

Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec

COP = \frac{16.97}{9.8} = 1.731

b)

COP ∝ \frac{1}{P_{in}}

P_{in} wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625

we know that

COP = \frac{heat\ obsorbed}{P_{in}}

P_{IN} = \frac{16.97}{35.62} = 0.4763

c) second law efficiency

\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%

exergy destruction os given as X = W_{IN} - X_{Q2}

                                                         = 9.8 - 0.473 = 9.3237 kW

8 0
2 years ago
1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bot
liberstina [14]

Answer:

T_{f} = 90.07998 ° C

Explanation:

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

                 Qc = -Qb

                M c_{e_Te} (T₁ -T_{f}) = m c_{e_air} (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

   ρ = m / V  

   V = 1000 -100 = 900 ml  

   V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³  

   V_air = 0.100 l = 0.1 10⁻³ m³  

Tea Mass  

     M = ρ V_te  

     M = 1100 0.9 10⁻³  

     M = 0.990 kg  

Air mass  

     m = ρ _air V_air  

     m = 1.225 0.1 10⁻³  

     m = 0.1225 10⁻³ kg  

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀  

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})  

Let's calculate  

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)  

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)  

T_{f} = 98097.11 / 1089.0002  

T_{f} = 90.07998 ° C  

This temperature decrease is very small and cannot be measured

3 0
2 years ago
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