Select the correct answerConsider the following SQL query: SELECT customer_id, MAX(total_amount) AS max_amount FROM Orders GROUP BY customer_id HAVING COUNT(*) > 1;What will be the output of this query?OptionsThe customer IDs and the maximum order amount for customers who have placed more than one order.An error because the HAVING clause is used without an aggregate function in the SELECT clause. An error because the GROUP BY clause is not properly used.The customer IDs and the average order amount for customers who have placed more than one order.

Select the correct answerWhat will be the result of the following query? SELECT customer_id, AVG(total_amount) AS avg_amount FROM Orders GROUP BY customer_id HAVING AVG(total_amount) > 200;OptionsReturns the customer IDs and average order amounts for all customers.Returns the customer IDs and total order amounts for customers with an average order amount greater than 200.Returns an error because the HAVING clause is not allowed in this context.Returns the customer IDs and average order amounts for customers with an average order amount greater than 200.

Which SQL query correctly finds the top 5 customers with the highest total order amounts?1.0 MarksSELECT CustomerID, COUNT(OrderAmount) AS TotalOrderAmount FROM Orders GROUP BY CustomerID ORDER BY TotalOrderAmount ASC LIMIT 5;SELECT CustomerID, SUM(OrderAmount) AS TotalOrderAmount FROM Orders GROUP BY CustomerID ORDER BY TotalOrderAmount DESC LIMIT 5;SELECT CustomerID, AVG(OrderAmount) AS TotalOrderAmount FROM Orders GROUP BY CustomerID ORDER BY TotalOrderAmount DESC LIMIT 5;SELECT CustomerID, SUM(OrderAmount) AS TotalOrderAmount FROM Orders WHERE OrderAmount > 100 GROUP BY CustomerID ORDER BY TotalOrderAmount DESC LIMIT 5;

29. What is the result of the following SQL Query? SELECT COUNT(*), AVG(amount) FROM Orders WHERE amount > 200; Number of orders with amount > 200 and the average amount of these orders.Total number of orders and the average amount of all orders.Number of customers with orders > 200 and the average order amount per customer.Total number of orders > 200 and the average number of orders.

We have following relationorders(order_id,customer_id,order_date,amount)1) Find out the number of orders for each customer by customer_id.2) Find out the total amount by order_id and order_date.3) Find out the number of orders for each customer by customer_id. Show onlycustomer_id with number of orders above 5

Query the customer_number from the orders table for the customer who has placed the largest number of orders.It is guaranteed that exactly one customer will have placed more orders than any other customer.The orders table is defined as follows:ColumnTypeorder_number (PK)intcustomer_numberintorder_datedaterequired_datedateshipped_datedatestatuschar(15)commentchar(200)Sample Inputorder_numbercustomer_numberorder_daterequired_dateshipped_date112017-04-092017-04-132017-04-12222017-04-152017-04-202017-04-18332017-04-162017-04-252017-04-20442017-04-182017-04-282017-04-25Sample Outputcustomer_number3ExplanationThe customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them only has one order.So the result is customer_number '3'. Follow up: What if more than one customer have the largest number of orders, can you find all the customer_number in this case?Optionsselect customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_number select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as elimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt group by customer_number) as eorder by e.cnt desclimit 1;