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Oksanka [162]
2 years ago
13

"Describe how implementation of a RAID Level 2 system would be beneficial to a university payroll system. In your own words, des

cribe the disadvantages of such a system, if any, in that environment, and if appropriate, suggest an alter-native RAID system and explain your reasoning"
Computers and Technology
1 answer:
Darina [25.2K]2 years ago
4 0

Answer:

In RAID 2 Hamming Code ECC Each piece of the information word is kept in touch with an information plate drive . Every datum word has its Hamming Code ECC word recorded on ECC circles. On Read, the ECC code confirms the right information or revises the single plate blunders.  

Strike Level 2 is one of the two innately equal mapping and assurance strategies characterized in the Berkeley paper. It has not been broadly sent in the business to a great extent since it requires an extraordinary plate highlights. Since plate creation volumes decides the cost, it is increasingly practical to utilize standard circles for the RAID frameworks.  

Points of interest: "On the fly" it gives information blunder amendment. Amazingly high information move rates is possible.Higher the information move rate required,better the proportion of information circles to ECC plates. Moderately basic controller configuration contrasted with the other RAID levels 3,4 and 5.  

Impediments: Very high proportion of the ECC plates to information circles with littler word sizes - wasteful. Passage level expense are extremely high - and requires exceptionally high exchange rate prerequisite to legitimize. Exchange rate is equivalent to that of the single plate, best case scenario (with shaft synchronization). No business executions can't/not industrially suitable.

You might be interested in
Consider the following relation:CAR_SALE(Car#, Date_sold, Salesperson#, Commission%, Discount_amt)Assume that a car may be sold
GalinKa [24]

Answer:

The answer to this question can be given as:

Normalization:

Normalization usually includes the division of a table into two or more tables as well as defining a relation between the table. It is also used to check the quality of the database design. In the normalization, we use three-level that are 1NF, 2NF, 3NF.

First Normal Form (1NF):

In the 1NF each table contains unique data. for example employee id.  

Second Normal Form (2NF):

In the 2NF form, every field in a table that is not a determiner of another field's contents must itself be a component of the table's other fields.

Third Normal Form (3NF):

In the 3NF form, no duplication of information is allowed.

Explanation:

The explanation of the question can be given as:

  • Since all attribute values are single atomic, the given relation CAR_SALE is in 1NF.  
  • Salesperson# → commission% …Given  Thus, it is not completely dependent on the primary key {Car#, Salesperson#}. Hence, it is not in 2 NF.                                                                                                        

The 2 NF decomposition:

        CAR_SALE_1(Car#, Salesperson#, Date_sold, Discount_amt)

         CAR_SALE_2(Salesperson#, Commission%)

  • The relationship in question is not in 3NF because the nature of a transitive dependence occurs
  • Discount_amt → Date_sold → (Car#, Salesperson#) . Thus, Date_sold is neither the key itself nor the Discount amt sub-set is a prime attribute.  

The 3 NF decomposition :

        CAR_SALES_1A(Car#, Salesperson#, Date_sold)

        CAR_SALES_1B(Date_sold, Discount_amt)

        CAR_SALE_3(Salesperson#, Commission%)

5 0
2 years ago
6. Write pseudo code that will perform the following. a) Read in 5 separate numbers. b) Calculate the average of the five number
Alex

Answer:

Pseudo CODE

a)

n= Input “Enter 5 integer value”

b)

sum=0.0

For loop with i ranging from 0 - 5

Inside loop sum=n[i]+sum

Outside loop avg= sum/5

Print avg

c)

small=n[0] # assume the first number in the list is smallest

large= n[0] # assume the first number in the list is largest

For loop with i ranging from 0 - 5

Inside loop if n[i]<small #if any another number is smaller than small(variable)

Inside if Then small=n[i]

Inside loop if n[i]>large # if any another number is larger than large(variable)

Inside if then large=n[i]

Print small

Print large

d)

print avg

print small

print large

8 0
2 years ago
What will be displayed after code corresponding to the following pseudocode is run? Main Set OldPrice = 100 Set SalePrice = 70 C
Tatiana [17]

Answer:

A jacket that originally costs $ 100 is on sale today for $ 80                                    

Explanation:

Main : From here the execution of the program begins

Set OldPrice = 100  -> This line assigns 100 to the OldPrice variable

Set SalePrice = 70   -> This line assigns 70to the SalePrice variable

Call BigSale(OldPrice, SalePrice)  -> This line calls BigSale method by passing OldPrice and SalePrice to that method

Write "A jacket that originally costs $ ", OldPrice  -> This line prints/displays the line: "A jacket that originally costs $ " with the resultant value in OldPrice variable that is 100

Write "is on sale today for $ ", SalePrice  -> This line prints/displays the line: "is on sale today for $ " with the resultant value in SalePrice variable that is 80

End Program -> the main program ends

Subprogram BigSale(Cost, Sale As Ref)  -> this is a definition of BigSale method which has two parameters i.e. Cost and Sale. Note that the Sale is declared as reference type

Set Sale = Cost * .80  -> This line multiplies the value of Cost with 0.80 and assigns the result to Sale variable

Set Cost = Cost + 20  -> This line adds 20 to the value of Cost  and assigns the result to Cost variable

End Subprogram  -> the method ends

This is the example of call by reference. So when the method BigSale is called in Main by reference by passing argument SalePrice to it, then this call copies the reference of SalePrice argument into formal parameter Sale. Inside BigSale method the reference &Sale is used to access actual argument i.e. SalePrice which is used in BigSale(OldPrice, SalePrice) call. So any changes made to value of Sale will affect the value of SalePrice

So when the method BigSale is called two arguments are passed to it OldPrice argument and SalePrice is passed by reference.

The value of OldPrice is 100 and SalePrice is 70

Now when method BigSale is called, the reference &Sale is used to access actual argument SalePrice = 70

In the body of this method there are two statements:

Sale = Cost * .80;

Cost = Cost + 20;

So when these statement execute:

Sale = 100 * 0.80 = 80

Cost = 100 + 20 = 120

Any changes made to value of Sale will affect the value of SalePrice as it is passed by reference. So when the Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice statement executes, the value of OldPrice remains 100 same as it does not affect this passed argument, but SalePrice was passed by reference so the changes made to &Sale by statement in method BigSale i.e.  Sale = Cost * .80; has changed the value of SalePrice from 70 to 80 because Sale = 100 * 0.80 = 80. So the output produced is:

A jacket that originally costs $ 100 is on sale today for $ 80                              

7 0
2 years ago
Assume the user types in 7 and 10. What is output by the following?
Marianna [84]

Answer:

Enter a number: 7

Enter a number: 10

Traceback (most recent call last):

 File "main.py", line 3, in <module>

   print (numi + num2)

NameError: name 'numi' is not defined

Explanation:

The typo in the print statement causes a run-time error, where obviously num1+num2 was expected, and an output of 17.

3 0
2 years ago
You are responsible for a rail convoy of goods consisting of several boxcars. You start the train and after a few minutes you re
maks197457 [2]

Answer:

The program in C++ is as follows:

#include <iostream>

#include <iomanip>

using namespace std;

int main(){

   int cars;

   cin>>cars;

   double weights[cars];

   double total = 0;

   for(int i = 0; i<cars;i++){

       cin>>weights[i];

       total+=weights[i];    }

   double avg = total/cars;

   for(int i = 0; i<cars;i++){

       cout<<fixed<<setprecision(1)<<avg-weights[i]<<endl;    }

   return 0;

}

Explanation:

This declares the number of cars as integers

   int cars;

This gets input for the number of cars

   cin>>cars;

This declares the weight of the cars as an array of double datatype

   double weights[cars];

This initializes the total weights to 0

   double total = 0;

This iterates through the number of cars

   for(int i = 0; i<cars;i++){

This gets input for each weight

       cin>>weights[i];

This adds up the total weight

       total+=weights[i];    }

This calculates the average weights

   double avg = total/cars;

This iterates through the number of cars

   for(int i = 0; i<cars;i++){

This prints how much weight to be added or subtracted

       cout<<fixed<<setprecision(1)<<avg-weights[i]<<endl;    }

4 0
1 year ago
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