N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million,

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Pat Oliver
Posts: 9
Joined: Sat Apr 18, 2009 4:42 am
Contact:

N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million,

Postby Pat Oliver » Sat Apr 18, 2009 5:33 am

I need some serious help! :confused:

The number N, in millions, of hearing impaired persons of age X can be approximated by N= -0.00006x^3 + 0.0006x^2 - 0.1x + 1.7. Approximate to the nearest million the number of hearing impaired persons ages 35 and 50.

User avatar
stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Sat Apr 18, 2009 12:42 pm

Pat Oliver wrote:The number N, in millions, of hearing impaired persons of age X can be approximated by N= -0.00006x^3 + 0.0006x^2 - 0.1x + 1.7. Approximate to the nearest million the number of hearing impaired persons ages 35 and 50.

To find the value, to the nearest million, of persons age x = 35, plug "35" in for "x". Round to the nearest whole number (since N is already in "millions").

To find the value, to the nearest million, of persons age x = 50, plug "50" in for "x". Round as above.

:wink:

Pat Oliver
Posts: 9
Joined: Sat Apr 18, 2009 4:42 am
Contact:

Re: N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million,

Postby Pat Oliver » Sun Apr 19, 2009 12:35 am

Thanks stapel_eliz! I overlooked something soooo simple. Thats why they say 2 heads are better than one! :)


Return to “Intermediate Algebra”

cron