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snow_tiger [21]
3 years ago
14

How long is the bachelor's program at Eth Zurich? 3 or 4 years?

Computers and Technology
1 answer:
natulia [17]3 years ago
4 0
The bachelor's program at Eth Zurich is 3 years.
I hope this helps!
:-)
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HACTEHA [7]

Answer:

The first one is Mobile Marketing Second Box is social media and the last is Search engine

Explanation:

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

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3 years ago
How many pictures can a 32 mb memory card hold?
nasty-shy [4]
This depends on the number of pixels are in each picture. hope this helps.
7 0
3 years ago
Cerinţa Se dau două numere naturale nenule n și p. Afișați în ordine crescătoare puterile lui n mai mici sau egale cu p. Date de
rjkz [21]

Given an array of integers (both odd and even), sort them in such a way that the first part of the array contains odd numbers sorted in descending order, rest portion contains even numbers sorted in ascending order.

Explanation:

  • Partition the input array such that all odd elements are moved to left and all even elements on right. This step takes O(n).
  • Once the array is partitioned, sort left and right parts individually. This step takes O(n Log n).

#include <bits/stdc++.h>  

using namespace std;  

void twoWaySort(int arr[], int n)  

{

   int l = 0, r = n - 1;  

   int k = 0;  

   while (l < r) {  

       while (arr[l] % 2 != 0) {  

           l++;  

           k++;  

       }

       while (arr[r] % 2 == 0 && l < r)  

           r--;  

       if (l < r)  

           swap(arr[l], arr[r]);  

   }  

   sort(arr, arr + k, greater<int>());  

   sort(arr + k, arr + n);  

}

int main()  

{  

   int arr[] = { 1, 3, 2, 7, 5, 4 };  

   int n = sizeof(arr) / sizeof(int);  

   twoWaySort(arr, n);  

   for (int i = 0; i < n; i++)  

       cout << arr[i] << " ";  

   return 0;  

}

8 0
3 years ago
i have a at&amp;t router and a 1000mbs Ethernet cable connecting from that to my net gear r7000 that can push +1000mbs. the cabl
Lady_Fox [76]
DSL ranges from 128Kbps to 3Mbps, so this would be your bottleneck.
7 0
3 years ago
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