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MA_775_DIABLO [31]
3 years ago
8

Write a program to find the product of 3 numbers

Computers and Technology
2 answers:
ziro4ka [17]3 years ago
6 0

Answer:

#include<iostream>

using namespace std;

int main()

{

   int a,b,c;

   cout<<"enter the value of a:";

   cin>>a;

   cout<<"enter the value of b:";

   cin>>b;

   cout<<"enter the value of c:";

   cin>>c;

   cout<<"product is:"<<(a*b*c);

   return 0;

}

Explanation:

AnnyKZ [126]3 years ago
6 0

The following codes have been written using Python.

x = int(input("Enter a number: "))

y = int(input("Enter a number: "))

z = int(input("Enter a number: "))

p = x*y*z

print(p, "is the product of", x, y, "and", str(z)+ ".")

Hope it helps. :)

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Read 2 more answers
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
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