I NEED HELP WITH #5 THROUGH #11 PLEASE: The following (simple) database has been

Question

I NEED HELP WITH #5 THROUGH #11 PLEASE:

The following (simple) database has been designed to keep track of customers and outstanding loans.Some sample data has been entered into the database. · Account: Accountlype Name Rate Terms Private Corporate Preferred 10.00% 7.00% 5.00% 30 60 90 Customer: CustomeriDFirstName 1123 1234 2134 2233 3324 LastName AccountType DateAcquired Barney Bugs Fred Herman George Jane Rubble Bunny Flintstone Munster Jetson HisWife 4/1/00 6 8 4/4/00 12/1/99 3/5/90 3/30/93 3/31/93 Loan: LoanlD 120 121 122 123 DistributionDate CustomerlD LoanAmount S500.00 S10,000.00 $500.00 $1,000.00 3/10/00 1234 4/4/00 2233 3/14/00 1234 3/15/00 1123 1. Using the above tables, show exactly what would be produced by the following SQL statement: SELECT LastName, DateAcquired FROM Customer WHERE Accounthpe = "A" ORDER BY LastNme; 2. Using the above tables, show exactly what would be produced by the following SQL statement: SELECT LastName, DateAcquired FROM Customer INNER JOIN Loan ON Customer·CustomerID = Loan·CustomerID WHERE AccountType = "A"

Explanation / Answer

#5. SELECT FirstName,LastName FROM Customer WHERE LastName LIKE 'M%' or LastName LIKE 'N%';
#6. select AccountType,count(CustomerID) as "Number Of Customer" from customer Group By AccountType;
#7. select sum(convert(trim(BOTH '$' FROM replace(LoanAmount,',','')),decimal)) as "Total Amount Of All Loan" from loan ;

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