Year f1(x)-52-0.531 Online video services such as YouTube have expanded according to the function N(t) =...

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Year f1(x)-52-0.531 Online video services such as YouTube have expanded according to the function N(t) = 52 t ^(0.531), where t represents the year that data was collected, starting with t = 1 corresponding to 2003. N is the number of online viewers in millions of people. -10 2003 1 2004 2 2005 3 2006 4 2007 5 2008 6 2009 7 2010 8 2011 9 2012 10 t Online video viewers (millions of people) 52 M 75.136 M 93.187 M 108.567 M (i.e., 52,000,000 people) (i.e., 75,136,000 people) (i.e., 93,187,000 people) (i.e., 108,567,000 people) (i.e., 122,224,000 people) 134.649 M (i.e., 134,649,000 people) 146.134 M (i.e., 146,134,000 people) ???.??? M 122.224 M 166.996 M (i.e., 166,996,000 people) 176.605 M (i.e., 176,605,000 people) How many million people viewed online videos in the year 2010? Type your answer as a number rounded to the nearest thousandth place (???.???). Year f1(x)-52-0.531 Online video services such as YouTube have expanded according to the function N(t) = 52 t ^(0.531), where t represents the year that data was collected, starting with t = 1 corresponding to 2003. N is the number of online viewers in millions of people. -10 2003 1 2004 2 2005 3 2006 4 2007 5 2008 6 2009 7 2010 8 2011 9 2012 10 t Online video viewers (millions of people) 52 M 75.136 M 93.187 M 108.567 M (i.e., 52,000,000 people) (i.e., 75,136,000 people) (i.e., 93,187,000 people) (i.e., 108,567,000 people) (i.e., 122,224,000 people) 134.649 M (i.e., 134,649,000 people) 146.134 M (i.e., 146,134,000 people) ???.??? M 122.224 M 166.996 M (i.e., 166,996,000 people) 176.605 M (i.e., 176,605,000 people) How many million people viewed online videos in the year 2010? Type your answer as a number rounded to the nearest thousandth place (???.???).

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The function Nt 52t0531 is an exponential function T... View the full answer