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aniked [119]
3 years ago
14

What set operator would you use if you wanted to see all of the Customers and the Employees including any records that may appea

r in both Customers and Employees? select firstname ,lastname ,phone from Customers select firstname ,lastname ,phone from Employees; Group of answer choices1 INTERSECT2 EXCEPT3 UNION ALL4 UNION
Computers and Technology
1 answer:
ikadub [295]3 years ago
7 0

Answer:

1. INTERSECT

Explanation:

Intersection is the set operator that gives us the information which is common to both tables. In this case the query will be:

SELECT firstname, lastname, phone FROM Customers

INTERSECT

SELECT firstname, lastname, phone FROM Employees;

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It identifies and removes viruses in computers
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4 years ago
_____ maintains consistency among data flow diagrams (dfds) by ensuring that input and output data flows align properly.
exis [7]

Indexing maintains consistency among data flow diagrams (dfds) by ensuring that input and output data flows align properly.

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
You have a mosaic dataset from your client. You want to use the mosaic dataset with other spatial data on the web. When you impo
sladkih [1.3K]

Answer:

Explanation:

Once a mosaic dataset as been created, the projection of the dataset cannot be changed or altered. Although the spatial reference section can be edited by going to mosaic database properties, then edit spatial reference section.

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4 years ago
Difference between sorting and filtering​
Serggg [28]
<h3>Sorting</h3>

The term “sorting” is used to refer to the process of arranging the data in <u>ascending or descending order</u>.

Example: Statistical data collected can be sorted alphabetically or numerically based on the value of the data.

<h3>Filtering</h3>

The process of data filtering involves selecting a <u>smaller part of your data set to view or analyze</u>. This is done by using that subset to view or analyze your data set as a whole.

Example: A complete set of data is kept, but only a portion of that set is used in the calculation, so the whole set is not used.

<em>Hope this helps :)</em>

5 0
1 year ago
Read 2 more answers
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