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Ymorist [56]
3 years ago
10

. Carly's Catering provides meals for parties and special events. In Chapter 2, you wrote an application that prompts the user f

or the number of guests attending an event, displays the company motto with a border, and then displays the price of the event and whether the event is a large one. Now modify the program so that the main() method contains only three executable statements that each call a method as follows:
Engineering
1 answer:
klio [65]3 years ago
6 0

Answer:

Explanation:

public class Event

  public final static int PRICE_PER_GUEST = 35;

  public final static int CUT_OFF = 50;

 

  //Attributes

  private String eventNum;

  private int noOfGuest;

  private int price;

 

  /**

  * param eventNum the eventNum to set

  */

  public void setEventNum(String eventNum)

      this.eventNum = eventNum;

 

 

  /**

  * param noOfGuest the noOfGuest to set

  */

  public void setNoOfGuest(int noOfGuest)

      this.noOfGuest = noOfGuest;

      this.price = this.noOfGuest * PRICE_PER_GUEST;

 

 

  /**

  * return the eventNum

  */

  public String getEventNum()

      return eventNum;

 

 

  /**

  * return the noOfGuest

  */

  public int getNoOfGuest()

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  1. To wear PPE
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  3. Know procedure to eliminate any threat
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Who's your favorite singer and WHT your favorite song​
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and my favorite song is popular loner

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Pls help me it’s due today
hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

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6 0
2 years ago
A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

                                                  = $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

            $=2 \pi (0.23)^2$

            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

     = 7043.42 N

3 0
2 years ago
Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb
Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0
3 years ago
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