Making sure

Making sure

Postby Guest » Thu Jan 09, 2020 1:58 pm

In 2000 the company had 300 customers. In 2015, this company already had
1200 customers. How many customers did the company have in 2008 if we know that
the number of customers growing is linear?


Just want to make sure, is 780 the correct answer?
Guest
 

Re: Making sure

Postby Guest » Sat Jan 11, 2020 11:57 am

Yes that is true.

300 + 8 * 60 = 780
Guest
 

Re: Making sure

Postby HallsofIvy » Sun Jan 12, 2020 11:15 pm

If growth is linear then we can write the formula as y= ax+ b, where x is the year and y is number if customers.

"In 2000 the company had 300 customers."
So when x= 2000, y= 300. 300= 2000a+ b.

"In 2015, this company already had 1200 customers."
So when x= 2015, y= 1200. 1200= 2015a+ b.

We can immediately eliminate "b" by subtracting the first equation from the second;
900= 15a so a= 900/15= 60. Then 300= 2000(60)+ b= 120000+ b so b= 300- 120000= -119700.
y= 60a- 119700.

"How many customers did the company have in 2008?"
y= 60(2008)- 119700= 120480- 119700= 780.

The company will had 780 clients in 2008.

(You can simplify the numbers by "coding" 2000 as 0, 20015 as 15, and 2018 as 18.)

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