The homepage is the page your browser displays when you first start the program

Select the correct answer.

irinina [24]

Answer:

A

Explanation:

6 0

3 years ago

Over the past week, every time Larry has started his computer, he has noticed that the time is not correct. Larry didn't conside

DiKsa [7]

Answer + Explanation:

You can tell Larry to set his device's location on 'enabled' at all times. That way, you get the time zone too. For example. if you live in the US, and your Location is not enabled, then the device may think that you are in China, or where the device was last at.

4 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 two characters do you use to tell the command that you are finished providing options and that the remaining data on the c

sergiy2304 [10]

Answer:

Two hyphen-minus characters (- -)

Explanation:

Using two hyphen-minus characters only, tend to recommend that the rest of the data are arguments and should not be treated as alternatives.

Hence, in this situation, the two characters a user will use to tell the command that a user is complete giving options and that the rest of the data on the command line is arguments is known as "two hyphen-minus characters (- -)"

8 0

3 years ago

Please write down a java program for below instruction.Min-Heap: Construct a Min_Heap from the following set of integers. Rememb

Leona [35]

Answer:

Following are the program in the Java Programming Language.

//define class

class Heap{

//set private integer data type array variable

private int a[];

//set two private integer data type variables

private int s,capacity;

//define private void function for down heap

private void downheap(int indx) {

//set if conditional statement

if(indx>=s)return;

//set integer type variable and initialize

int lft=2*indx+1;

int rgt=2*indx+2;

int min=indx;

//set the if-else if conditional statement

if(rgt<=s&&a[rgt]<a[indx])

min=rgt;

else if(lft<=s&&a[lft]<a[min])

min=lft;

//check index is not equal to min

if(indx!=min){

//perform swapping

int temp=a[indx];

a[indx]=a[min];

a[min]=temp;

downheap(min);

}

//define private void type function for upheap

private void upheap(int indx) {

if(indx==0)return;

int parent=(indx-1)/2;

if(a[indx]<a[parent]) {

int temp=a[indx];

a[indx]=a[parent];

a[parent]=temp;

upheap(parent);

}

public Heap(int q) {

a=new int [q];

s=0;

capacity=q;

}

public int _size () {

return s;

}

public int minLookup() {

if(s>0)return a[0];

return 0;

}

public int remove() {

int data=a[0];

a[0]=a[s-1];

s--;

downheap(0);

return data;

}

public void add (int data) {

if(s>=capacity)return;

a[s++] = data;

upheap(s-1);

}

public void print_heap() {

System.out.print("Heap : ");

for(int i=0;i<s;i++) {

System.out.print(a[i]+" ");

}

System.out.println();

}

//define class to define main method

public class Main {

//main method

public static void main(String srgs[]) {

//cretae class object

Heap heap = new Heap(20);

//call function through object with argument list

heap.add(1);

heap.add(19);

heap.add(7);

heap.add(5);

heap.add(6);

heap.add(42);

heap.add(21);

heap.add(13);

heap.add(56);

heap.add(78);

heap.add(29);

heap.add(3);

heap.add(14);

heap.print_heap();

//print the Output and remove the minimum element

System.out.println("Minimum element : "+heap.remove());

heap.print_heap();

System.out.println("minimum element : "+heap.remove());

heap.add(45);

heap.print_heap();

}

<u>Output:</u>

Heap : 1 5 3 13 6 7 21 19 56 78 29 42 14

Minimum element : 1

Heap : 3 5 7 13 6 14 21 19 56 78 29 42

Minimum element : 3

Heap : 7 5 21 13 6 14 42 19 56 78 29 45

Explanation:

Here, we define a class "Heap" inside the class.

Set private integer data type array variable.
set private two integer data type variables.
Then, we define private void type function "downheap()" for down heap and pass an integer data type argument list in its parameter "indx".

Then, we define private void type function "upheap()" for upper heap and pass an integer data type argument list in its parameter "indx".
Define constructor of the class for the capacity pf the array list.
Define void type function "add()" for adding elements in heap.
Define void type function "print_heap()" to print the heap.

Finally, we define class to define main function and we create the object of the class "Heap" then, call all the functions through the class object abd print the output.

4 0

3 years ago