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julia-pushkina [17]
3 years ago
10

The following table describes the required fields for two classes and typical values stored in those fields.

Computers and Technology
1 answer:
enot [183]3 years ago
8 0

Answer:

hope this helps,if it did pls mark my ans as brainliest

Explanation:

using System;

class Printer{ string companyName; int pagesOutput;

double price;

public Printer(string companyName,int pagesOutput,double price){

this.companyName=companyName; this.pagesOutput=pagesOutput; this.price=price;

}

public virtual void Display(){ Console.WriteLine("companyName: "+companyName+" pagesOutput: "+pagesOutput+" price: "+price);

}

}

class PrinterCumScanner:Printer{ int imageDpi;

public PrinterCumScanner(string companyName,int pagesOutput,double price,int imageDpi):base(companyName,pagesOutput,price){ this.imageDpi=imageDpi; }

public override void Display(){ base.Display(); Console.WriteLine("imageDpi: "+imageDpi);

}

}

public class Program { static void Main(string[] args) { const int NUM_PRINTERS=3;

Printer []stock=new Printer[NUM_PRINTERS];

stock[0]=new Printer("HP",40,89.50);

stock[1]=new PrinterCumScanner("HP",40,215.0,320); stock[2]=new PrinterCumScanner("Cannon",30,175.0,240);

foreach(Printer aPrinter in stock){ aPrinter.Display();

}

}

}

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Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has0.020 defects /cm2Assume a 20 cm diameter wafer has a c
Vitek1552 [10]

Answer:

1. yield_1=0.959 and yield_2=0.909

2. Cost_1=0.148 and Cost_2=0.165

3. New area per die=1.912 cm^2 and yield_1=0.957

   New area per die=2.85 cm^2  and yield_2=0.905

4. defects=0.042 per cm^2 and defects=0.026 per cm^2

Explanation:

1. Find the yield for both wafers.

yield= 1/(1+(defects per unit area*dies per unit area/2))^2

Wafer 1:

Radius=Diameter/2=15/2=7.5 cm

Total Area=pi*r^2=pi(7.5)^2=176.71 cm^2

Area per die= 176.71/84=2.1 cm^2

yield_1= 1/(1+(0.020*2.1/2))^2

yield_1=1/1.04244=0.959

Wafer 2:

Radius=Diameter/2=20/2=10 cm

Total Area=pi*r^2=pi(10)^2=314.159 cm^2

Area per die= 314.159/100=3.14 cm^2

yield_2= 1/(1+(0.031*3.14/2))^2

yield_2=1/1.0997=0.909

2. Find the cost per die for both wafers.

Cost per die= cost per wafer/Dies per wafer*yield

Wafer 1:

Cost_1=12/84*0.959=0.148

Wafer 2:

Cost_2=15/100*0.909=0.165

3. If the number of dies per wafer is increased by 10% and the defects per area unit increases by 15%, find the die area and yield.

Wafer 1:

There is a 10% increase in the number of dies

10% of 84 =8.4

New number of dies=84.4+8=92.4

There is a 15% increase in the defects per cm^2

15% of 0.020=0.003

New defects per area= 0.020 + 0.003=0.023 defects per cm^2

New area per die= 176.71/92.4=1.912 cm^2

yield_1= 1/(1+(0.023*1.912/2))^2=0.957

Wafer 2:

There is a 10% increase in the number of dies

10% of 100=10

New number of dies=100+10=110

There is a 15% increase in the defects per cm^2

15% of 0.031=0.0046

New defects per area= 0.031 + 0.00465=0.0356 defects per cm^2

New area per die= 314.159/110=2.85 cm^2

yield_2= 1/(1+(0.0356*2.85/2))^2=0.905

4. Assume a fabrication process improves the yield from 0.92 to 0.95. Find the defects per area unit for each version of the technology given a die area of

Assuming a die area of 2cm^2

We have to find the defects per unit area for a yield of 0.92 and 0.95

Rearranging the yield equation,

yield= 1/(1+(defects*die area/2))^2

defects=2*(1/sqrt(yield) - 1)/die area

For 0.92 technology

defects=2*(1/sqrt(0.92) - 1)/2

defects=0.042 per cm^2

For 0.95 technology

defects=2*(1/sqrt(0.95) - 1)/2

defects=0.026 per cm^2

6 0
3 years ago
What is civil construction?
sasho [114]
Civil construction is any construction in a society that belongs to the group of core infrastructure such as <span>pipelines, telecommunications, sewers, water treatment systems, highways, roads, bridges etc.</span>
6 0
3 years ago
In a shallow copy, ____. the values of two or more pointers are copied into a list two or more pointers of the same type point t
strojnjashka [21]

Answer:

Explanation:

A Shallow copy is a bit-wise (unit of data) copy of an object that can be found/used in an object-oriented programming language code, the copy contains all of the same exact values that exist within the original object. In a shallow copy, two or more pointers of the same type point to the same memory location.

8 0
3 years ago
Read 2 more answers
The cell tower’s base transmission station is connected to a mobile ____ station.
geniusboy [140]

The cell tower’s base transmission station is connected to a mobile switching station.

<h3>What is a switching station?</h3>

A switching station is known to be a kind of a substation that is known to be without transformers and it is one that often operate  only based or at a single voltage level.

Note that Switching stations are known to be often used as collector and they also acts as a distribution stations.

Hence,  The cell tower’s base transmission station is connected to a mobile switching station.

Learn more about transmission station from

brainly.com/question/1312500

#SPJ1

8 0
2 years ago
You are the network administrator for a growing business. when you were initially hired, the organization was small and only a s
k0ka [10]
Usually, network devices can be set up to send their syslog messages to a syslog server, so you'd just need a *nix computer with syslog setup to receive messages via the network, and you can read stuff there. You also get the advantage of having reglar expressions so that "noise" (messages you don't care about) isn't displayed.

Network devices can also be set up to use SNMP (Simple Network Management Protocol).

6 0
3 years ago
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