Consider the following Relation Coffee with two columns NAME and PRICE, COFFEE

NAME

PRICE

Espresso

1000

Cappuccino

900

Espresso Macchiato

1500

Americano

500

Cafe Breve

600

Tea

NULL


SELECT Count (∗) FROM COFFEE

WHERE PRICE > ANY (SELECT PRICE FROM COFFEE)

output is

1
6
2
4
3
5
4
3

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