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trapecia [35]
3 years ago
15

Whats the best way to get into a computers programming to get past firewalls

Computers and Technology
2 answers:
aniked [119]3 years ago
8 0

well you would need a hack code to start with.

raketka [301]3 years ago
6 0

you can use the software psiphon

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The software that requests mail delivery from the mail server to an Internet device is known as mail ____ software.
OleMash [197]

The software that requests mail delivery from the mail server to an Internet device is known as mail client software.

Explanation:

Mail client software is a form of communication in which electronic messages are created and transferred between two or multiple devices connected to a network series or stream.

8 0
3 years ago
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 are the differences between a centralized and distributed database architecture
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Answer:

The difference between the centralized and the distributed database architecture are as follows:

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  • The centralized database refers to the proper planning and the helps in the decision making in the management system. Whereas, in the distributed database system the data are mainly distributed in the multiple files and the user can easily access the near by file in the database system.
  • In the centralized database, if the database are get fail then the overall system come to halt. On the other hand in the distributed system, if the component get failed then the performance of the system are get reduced.

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Select the correct answer. Adrian sent his manager Stacy an email message that required her immediate attention. Stacy was unabl
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