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.
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N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million, 
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N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million,
by Pat Oliver on Sat Apr 18, 2009 5:33 am
I need some serious help!
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.
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.
- Pat Oliver
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by stapel_eliz on 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.
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stapel_eliz - Posts: 1701
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Re: N = 00.00006x^3 + 0.006x^2 - 0.1x + 1.7. To nearest million,
by Pat Oliver on 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! 
- Pat Oliver
- Posts: 9
- Joined: Sat Apr 18, 2009 4:42 am
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