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2 years ago

Which of the following statements is not true of web storage?

Computers and Technology

1 answer:

slavikrds [6]2 years ago

3 0

Answer:

The data in web storage is passed to the server with every HTTP request.

Explanation:

There are basically two Web storage APIs Session storage and local storage.Both can store data up to 5MB. They are supported by every modern browser.You can store data in local storage indefinitely and for browser session in session storage.There is no data or information in HTTP request header.So we conclude that option 4 is the answer.

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Andy works for a TV broadcasting company. He needs to set up a network covering a small area on the work floor. However, he noti

katen-ka-za [31]

I believe it’s B)LAN

Explanation:
A LAN connects computers over a relatively short distance, allowing them to share data, files, and resources. For example, a LAN may connect all the computers in an office building, school, or hospital. Typically, LANs are privately owned and managed.

4 0

2 years ago

The inflationary gap occurs when you obtain no increase in output, but only an increase in the Average Price Level from an incre

Katarina [22]

<span>B. Second phase of the Keynesian LRAS Curve.</span>

8 0

3 years ago

Here's something to stop you from getting repetitive when writing essays. Write a program that reads multiple lines of plain tex

Marianna [84]

Answer:

The output is attached below

Explanation:

d = {}

while True:

line = input("Enter line: ")

if len(line)==0:

break

token = line.split(' ')

for var in token:

try:

if len(var)==0:

continue

count = d[var]

d[var] = count + 1

except KeyError:

d[var] = 1

pass

for word in sorted(d):

print(word , d[word])

------------------

7 0

3 years ago

Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi

Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

5 0

3 years ago

Create the following SQL Server queries that access the Northwind database. Place them in a zip file and place the zip file in t

neonofarm [45]

Answer:

1. SELECT e.EmployeeID, e.FirstName, e.LastName, COUNT(*) as OrderByEmployee FROM Employees e

JOIN Orders o

ON e.EmployeeID = o.EmployeeID

WHERE YEAR(o.OrderDate) = '1996'

GROUP BY e.EmployeeID,e.FirstName,e.LastName

ORDER BY OrderByEmployee DESC

2. SELECT o.OrderID,c.CustomerID, o.OrderDate,SUM((od.Quantity)*

od.UnitPrice - od.Discount)) as TotalCost FROM Orders o

JOIN [Order Details] od

ON od.OrderID = o.OrderID

JOIN Customers c

ON o.CustomerID = c.CustomerID

WHERE (MONTH(OrderDate)= 7 OR MONTH(OrderDate) = 8) and

YEAR(OrderDate) = 1996

GROUP BY o.OrderID,c.CustomerID, o.OrderDate

ORDER BY TotalCost DESC

3. SELECT c.CustomerID, c.CompanyName, c.City, COUNT(o.OrderID) as TotalOrder, SUM(od.Quantity* od.UnitPrice) as TotalOrderAmount FROM Customers c

JOIN Orders o

ON o.CustomerID = c.CustomerID

JOIN [Order Details] od

ON od.OrderID = o.OrderID

WHERE c.Country = 'USA'

GROUP BY c.CustomerID, c.CompanyName, c.City

ORDER BY c.City, c.CustomerID

4. SELECT c.CustomerID, c.CompanyName, c.City, COUNT(o.OrderID) as TotalOrder, SUM(od.Quantity* od.UnitPrice) as TotalOrderAmount FROM Customers c

JOIN Orders o

ON o.CustomerID = c.CustomerID

JOIN [Order Details] od

ON od.OrderID = o.OrderID

WHERE c.Country = 'USA'

GROUP BY c.CustomerID, c.CompanyName, c.City

HAVING COUNT(o.OrderID) > 3

ORDER BY c.City, c.CustomerID

5. SELECT p.ProductID, p.ProductName, s.ContactName as SupplierName, MAX(o.OrderDate) as LastOrderDateOfProduct FROM Products p

JOIN Categories c

ON c.CategoryID = p.CategoryID

JOIN Suppliers s

ON s.SupplierID = p.SupplierID

JOIN [Order Details] od

ON od.ProductID = p.ProductID

JOIN Orders o

ON o.OrderID = od.OrderID

where c.CategoryName = 'Grains/Cereals'

GROUP BY p.ProductID, p.ProductName, s.ContactName

ORDER BY SupplierName, p.ProductName

6. SELECT p.ProductID, p.ProductName, Count(DISTINCT c.CustomerID ) as OrderByThisManyDistinctCustomer

FROM Products p

JOIN [Order Details] od

ON od.ProductID = p.ProductID

JOIN Orders o

ON o.OrderID = od.OrderID

JOIN Customers c

ON c.CustomerID = o.CustomerID

where YEAR(o.OrderDate) = 1996

GROUP BY p.ProductID, p.ProductName

Explanation:

The six query statements returns data from the SQL server Northwind database. Note that all the return data are grouped together, this is done with the 'GROUP BY' Sql clause.

8 0

3 years ago