Answer:
SELECT Count(order_invoice) as number_of_invoices, Max(order_invoice) as maximum_invoice, Min(order_invoice) as minimum_invoice, Avg(order_invoice) as average_invoice
FROM vendor JOIN invoice ON invoice.id = vendor.id
WHERE order_invoice > 1
ORDER BY number_of_invoices DESC
Explanation:
The select statement of the SQL or structured query language returns twelve rows of four columns from the inner join of the vendor and invoice table in a database where the order_invoice column in the invoice table is greater than one. The result of the query is ordered by the alias column "number_of_invoices" in descending order.
Answer:
We use SQL Not Equal comparison operator (<>) to compare two expressions. For example, 10<>11 comparison operation uses SQL Not Equal operator (<>) between two expressions 10 and 11
Explanation:
a. speed + 12 - miles * 2 = 10 + 12 - 5 * 2. With order of operations, we do the multiplication first so the equation is now 10 + 12 - 10 = 22 - 10 = 12
b. speed + miles * 3 = 10 + 5 * 3 and again, order of operations gives us 10 + 15 = 25
c. (speed + miles) * 3 = (10 + 5) * 3 = 15 * 3 = 45
d. speed + speed * miles + miles = 10 + 10 * 5 + 5 = 10 + 50 + 5 = 60 + 5 = 65
e. (10 – speed) + miles / miles = (10 - 10) + 5 / 5 = 0 + 5 / 5 = 5 / 5 = 1
Answer: 10 students
Explanation:
Students younger than 10 = 50%
Students that are 10years old = 1/20 = 1/20 × 100 = 5%
Students that are older than 10 but younger than 12 = 1/10= 1/10 × 100 = 10%
Students that are 12 years or older
= 100% - (50% + 5% + 10%)
= 100% - 65%
= 35%
This means that 35% of the students are 12 years or older and we've been given the number as 70.
Let's say the total number of students is x. Therefore,
35% of x = 70
0.35 × x = 70
0.35x = 70
x = 70/0.35
x = 200
The total number of students is 200.
Therefore, the number of students that are 10years will be:
= 1/20 × 200
= 10 students
Therefore, 10 students are 10 years.
