document.write( "Question 92167: The formula b(t) = 1.353(1.9025)^t represents the number of broadband internet users (in millions) where t if the time in years measured from 1998. How many broadband users (to the nearest million) does this formula predict for 2004? \n" ); document.write( "
Algebra.Com's Answer #67004 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Given:
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\n" ); document.write( "\"b%28t%29+=+1.353%281.9025%29%5Et\"
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\n" ); document.write( "In 2004 six years will have passed since the time in years measured from 1998. So you need
\n" ); document.write( "to define t as being 6 for this problem. That makes the problem become:
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\n" ); document.write( "\"b%28t%29+=+1.353%281.9025%29%5E6\"
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\n" ); document.write( "Raising 1.9025 to the sixth power can be done directly on a scientific calculator or with
\n" ); document.write( "a regular calculator you just multiply:
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\n" ); document.write( "1.9025*1.9025*1.9025*1.9025*1.9025*1.9025
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\n" ); document.write( "The answer to this part of the problem is that 1.9025 raised to the sixth power is 47.41851975
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\n" ); document.write( "Substituting this results in the problem becoming:
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\n" ); document.write( "\"b%28t%29+=+1.353%2A47.41851975\"
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\n" ); document.write( "and doing the multiplication results in the final answer of:
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\n" ); document.write( "\"b%28t%29+=+64.15725723\"
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\n" ); document.write( "So the answer is that there should have been 64.15725723 million broadband users in 2004.
\n" ); document.write( "Rounding this to the nearest million as specified by the problem reduces the answer to
\n" ); document.write( "64 million.
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\n" ); document.write( "Hope this helps you to understand the problem.
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