Question

the number of customers for a new online business can be modeled by y=10x^2+50x+300 where x represents the number of months since the business started which is the best prediction for the number of customers in month 20?

Answer 1

5300 Customers.

Work: 10(20)^2+50(20)+300

We are changing x to 20 because x represents the number of months and it asks for the amount of customers on month 20.

Answer 2

5300 no. of customers

Explanation:

Given :
`y=10x^2+50x+300` model of new online business

where x represents the number of months

To find: Best prediction for the number of customers in month 20

Solution: The model is given for x months =
`y=10x^2+50x+300`

The model is given for 20 months=
`y=10(20)^2+50(20)+300`

⇒
`y=10(400)+1000+300`

⇒
`y=10(400)+1000+300`

⇒
`y=5300`

Therefore, 5300 no. of customers is predicted