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Lostsunrise [7]
1 year ago
9

Which layer of the OSI model do network administrators typically check first when issues arise?

Computers and Technology
1 answer:
bazaltina [42]1 year ago
4 0

Answer:

B

Explanation:

Typically I check the physical layer to ensure that the nodes are plugged accordingly.

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Answer:

ios, aNDROID, BLUETOOTH , AND BLACK BERRY

Explanation:

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If a user wants to change one small section of the formatting of a document and leave the rest the same, which
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Evaluating how current, credible, and unbiased a source is ensures:
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The answer is

A. your information is more accurate and useful

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2 years ago
Write a function called play_round that simulates two people drawing cards and comparing their values. High card wins. In the ca
elena-s [515]

Answer:

Here you go, Change it however you'd like :)

Explanation:

import random as r

def play_round(p1, p2):

   cards = [1,2,3,4,5,6,7,8,9,10,"J","Q","K","A"]

   play1 = r.choice(cards)

   play2 = r.choice(cards)

   

   while play1 == play2:

       play1 = r.choice(cards)

       play2 = r.choice(cards)

   

   if cards.index(play1) > cards.index(play2):

       return f"{p1}'s Card: {play1}\n{p2}'s Card: {play2}\nThe Winner is {p1}"

   else:

       return f"{p1}'s Card: {play1}\n{p2}'s Card: {play2}\nThe Winner is {p2}"

print(play_round("Bob","Joe"))

8 0
2 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
2 years ago
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