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3 years ago

Media and social control

Computers and Technology

1 answer:

Fofino [41]3 years ago

5 0

Answer:

Media:

the main means of mass communication (broadcasting, publishing, and the Internet) regarded collectively.

Social control:

Social control is a concept within the disciplines of the social sciences.

Social control is the study of the mechanisms, in the form of patterns of pressure, through which society maintains social order and cohesion. ... Regardless of its source, the goal of social control is to maintain conformity to established norms and rules.

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In 5-10 sentences, describe the purposes, design considerations, and common elements of most brochures.

lozanna [386]

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3 years ago

What is a reason to use a pseudo-class in Dreamweaver?

gulaghasi [49]

Answer: Option C is correct

Explanation:

Option C is correct

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3 years ago

Which statement is FALSE? If a method does not return a value, the return-value-type in the method declaration can be omitted. P

Tcecarenko [31]

Answer:

If a method does not return a value, the return-value-type in the method declaration can be omitted.

Explanation:

If a method does not return a value, the return-value-type in the method declaration can be omitted is a false statement.

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3 years ago

Consider the following statement, which is intended to create an ArrayList named a to store only elements of type Thing. Assume

bogdanovich [222]

Answer:

B: new ArrayList()

Explanation:

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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