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Alexandra [31]
2 years ago
13

What is the purpose of secondary memory?

Computers and Technology
1 answer:
grigory [225]2 years ago
8 0

Answer:

secondary memory is usually used as mass storage.

You might be interested in
Which of the following files has the Ionic styles? ionic.bundle.css ionic.js ionic.css ionic.bundle.js
Law Incorporation [45]

Answer:

The correct option is ionic.css

Explanation:

Ionic style uses modes to customize the look of components. Each platform has a default mode, but this can be overridden.

And of all the options, only the ionic.css possesses the ionic styles

4 0
4 years ago
Var amount = 100;
IRISSAK [1]

Answer:

0.1

Explanation:

The value of amount is 100.So in the if-else-if ladder first condition in if will be checked since the amount is greater than 50 so the value of discount will become 0.1 and the execution of if else if ladder will be finish.The compiler will not be executing elseif or else.So the answer to this question is 0.1.

4 0
3 years ago
QUESTION 7
marishachu [46]
False, it needs to be properly cited because that is plagiarism which is illegal
6 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
Which XXX declares a student's name. public class Student { XXX private double myGPA; private int myID; public int getID() { ret
maw [93]
<h2>Question:</h2>

<em>Which XXX declares a student's name. </em>

public class Student {

   XXX

   private double myGPA;  

   private int myID;

   public int getID() {

        return myID;

   }

}

<em>Group of answer choices </em>

a. String myName;

b. public String myName;

c. private myName;

d. private String myName;

<h2>Answer:</h2>

private String myName;

<h2>Explanation:</h2>

To declare a student's name, the following should be noted.

i. The name of the student is of type <em>String</em>

ii. Since all of the instance variables (myGPA and myID) have a <em>private</em> access modifier, then myName (which is the variable name used to declare the student's name) should be no exception. In other words, the student's name should also have a <em>private</em> access.

Therefore, XXX which declares a student's name should be written as

<em>private String myName;</em>

<em></em>

Option (a) would have been a correct option if it had the <em>private</em> keyword

Option (b) is not the correct option because it has a <em>public</em> access rather than a <em>private</em> access.

Option (c) is not a valid syntax since, although it has a <em>private</em> access, the data type of the variable <em>myName</em> is not specified.

Option (d) is the correct option.

4 0
3 years ago
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