Answer:
8.28%
Step-by-step explanation:
According to the U.S. Census Bureau, the legal official department in charge of the National Census every ten years, the population of the U.S.A in 2010 was 308,745,538 inhabitants.
If about 25,585,000 had diabetes, then cross multiplying
308,745,538-------100%
25,585,000---------x%
and
[tex]\frac{308,745,538}{25,585,000}=\frac{100}{x}[/tex]
and so, about
[tex]x=\frac{25,585,000*100}{308,745,538}=8.28 \%[/tex]
Americans had diabetes in 2010.
