Question 841582
What they want you to do is find the slope of
a straight line that passes through the points
( 71.1, 0 ) and ( 73.4, 5 )
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{{{ E(t) }}} is plotted on the vertical axis
{{{ t }}} is plotted on the horizontal axis
I call the year 1992 {{{ t = 0 }}}
Then 1997 becomes {{{  t = 5 }}} ( 5 years later )
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Use the general point-slope formula
{{{ ( E - 71.1 ) / ( t - 0 ) = ( 73.4 - 71.1 ) / ( 5 - 0 ) }}}
{{{ ( E - 71.1 ) / t = ( 73.4 - 71.1 ) / 5 }}}
Multiply both sides by {{{ 5t }}}
{{{ 5*( E - 71.1 ) = t*( 73.4 - 71.1 ) }}}
{{{ 5E - 355.5  = 2.3t }}}
{{{ 5E = 2.3t + 355.5 }}}
{{{ E = .46t + 71.1 }}}
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Check the answer:
( 73.4, 5 )
{{{ E = .46t + 71.1 }}}
{{{ 73.4 = .46*5 + 71.1 }}}
{{{ 73.4 = 2.3 + 71.1 }}}
{{{ 73.4 = 73.4 }}}
OK
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2003 - 1992 = 11 years
{{{ t = 11 }}}
{{{ E = .46t + 71.1 }}}
{{{ E = .46*11 + 71.1 }}}
{{{ E = 5.06 + 71.1 }}}
{{{ E = 76.16 }}}
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{{{ E(11) = 76.16 }}} also
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Here's a plot of the line:
{{{ graph( 400, 400, -2, 15, -10, 100, .46x + 71.1 ) }}}