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SVEN [57.7K]
3 years ago
10

An organization is planning to implement a VPN. They want to ensure that after a VPN client connects to the VPN server, all traf

fic from the VPN client is encrypted. Which of the following would BEST meet this goal?a. Split tunnelb. Full tunnelc. IPsec using Tunnel moded. IPsec using Transport mode
Computers and Technology
1 answer:
Sonja [21]3 years ago
7 0

Answer:

B. Full tunnel.

Explanation:

In this scenario, an organization is planning to implement a virtual private network (VPN). They want to ensure that after a VPN client connects to the VPN server, all traffic from the VPN client is encrypted. The best method to meet this goal is to use a full tunnel.

A full tunnel is a type of virtual private network that routes and encrypts all traffics or request on a particular network. It is a bidirectional form of encrypting all traffics in a network.

On the other hand, a split tunnel is a type of virtual private network that only encrypts traffic from the internet to the VPN client.

Hence, the full tunnel method is the most secured type of virtual private network.

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larisa [96]

Answer:

pp popo

Explanation:

6 0
3 years ago
Read 2 more answers
Suppose Pentax develops a new photo editing software and negotiates agreements with a Vietnamese manufacturer to assemble most P
Elis [28]

Answer:

Contract manufacturing

Explanation:

Contract manufacturing involves outsourcing a portion of a manufacturing process of a product to another company.

5 0
3 years ago
For the given program, how many print statements will execute? public static void printShippingCharge(double weight) { if((weigh
Step2247 [10]

CORRECT QUESTION:

For the given program, how many print statements will execute?

public static void printShippingCharge(double weight) { if((weight > 0.0) && (weight <= 10.0)){ System.out.println(weight * 0.75); }

else if((weight > 10.0) && (weight <= 15.0)) { System.out.println(weight * 0.85); }

else if((weight > 15.0) && (weight <= 20.0)) { System.out.println(weight * 0.95); } }

public static void main(String args[]) {

printShippingCharge(18);

printShippingCharge(6);

printShippingCharge(25); }

Answer:

Two of the print statements will output values

Explanation:

These two calls to the printShippingCharge are the ones that will output values: printShippingCharge(18); and printShippingCharge(6);

The first if statement is true when the value of weight is 6 (weight > 0.0) && (weight <= 10.0)

The Third if statement  is true when weight is 18 (weight > 15.0) && (weight <= 20.0)

NOTE: That I made correction to the variable weight. The question isn't consistent with the variable name. It used weight and itemWeight at different points this will lead to a compiller error

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
The idea of supply and demand is based on the development of
Crazy boy [7]

Answer:

The idea of supply and demand is based on the development of sufficient products to meet consumer wants.

Explanation:

In economic theory supply and demand explains the relation between price ,sellers and buyers of a resource.

If the demand increases and supply increases it dose not have any affect on price.

If demand increase and supply decrease it triggers an increase in price

If demand decrease and supply increase it triggers decrease in price.

8 0
3 years ago
Read 2 more answers
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