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help me solve this math problem?

there's this bar graph that shows the number of American without health insurance from 2000 through 2003 :

in 2000 the number of people without health insurance is 39.8 million

in 2001 the number of people without health insurance is 41.3 million

in 2002 the number of people without health insurance is 43.4 million

in 2003 the number of people without health insurance is 44.9 million

in 2000, there was 39.8 millions Americans without health insurance. this number increased at an average rate of 1.7 million people per year. use this description to solve ;

the question :

determine when the number of Americans without health insurance will exceed the number in 2003 by 8.5 million

how do i solve? where do i begin ?

2 Answers

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  • 1 decade ago
    Favourite answer

    They tell you the average rate of increase per year is 1.7 million, so you do not even need to work it out [ (44.9 - 39.8) / 3 = 1.7 ]. Use this to work out how many years for an increase of 8.5 million ==>

    8.5 / 1.7 = 5 years, i.e. 2008 .... answer.

    Source(s): experience
  • 4 years ago

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    RE :Help me solve this math problem?

    there's this bar graph that shows the number of American without health insurance from 2000 through 2003 :

    in 2000 the number of people without health insurance is 39.8 million

    in 2001 the number of people without health insurance is 41.3 million

    in 2002 the number of people without health insurance is 43.4 million

    in 2003 the number of people without health insurance is 44.9 million

    in 2000, there was 39.8 millions Americans without health insurance. this number increased at an average rate of 1.7 million people per year. use this description to solve ;

    the question :

    determine when the number of Americans without health insurance will exceed the number in 2003 by 8.5 million

    how do i solve? where do i begin ?

    Follow 1 answer

    Source(s): I recommend that you visit this website where onel can compare quotes from different companies: http://coverage-finder.net/index.html?src=5YAwkrcu...
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