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ki77a [65]
2 years ago
6

There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T

he attendant opens the booth and improves the service rate over time following the function u(t) = 1.1 + .30t, where u(t) is in vehicles per minute and t is in minutes. I need to find the maximum queue length. I included my solution below for the total delay.
u(t) = 4 when t = (4-1.1)/0.3 = 9.67mins
(9.67 to t) ? (1.1+0.3t)dt = 10
? 1.1t + 0.15t2 ](9.67 to t)= 10
? t = 11.97mins
So, delay = 11.97 – 9.67 = 2.30mins
Engineering
1 answer:
dmitriy555 [2]2 years ago
3 0

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

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ivolga24 [154]

Answer:

A) strength decreases, chemical resistance decreases, and thermal insulation increases

Explanation:

Strength always decreases, chemical resistence decreases, and thermal condictivity must be reduced therefore themal insulation must increase.

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AVprozaik [17]

Answer:

The correct option is;

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Explanation:

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8 0
2 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
2 years ago
The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-
jarptica [38.1K]

Answer:

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Explanation:

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Allowable Shear Stress = Torque x Radius / pi/2 x radius^4

Putting the values we have;

T = 2000 N/m

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Putting values in formula;

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5 0
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kupik [55]

Answer:

month = input("Input the month (e.g. January, February etc.): ")

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season = 'spring'

elif month in ('July', 'August', 'September'):

season = 'summer'

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print("Season is",season)

Explanation:

4 0
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