Question 1006197
{{{ P(t) = 70000*( 1 - e^(-.0009t) ) }}}
(a)
When is the ratio {{{ P(t) / 70000 = .1 }}} ?
{{{ P(t) = .1*70000 }}}
{{{ P(t) = 7000 }}}
{{{ 7000 = 70000*( 1 - e^(-.0009t) ) }}}
{{{ .1 = 1 - e^( -.0009t ) }}}
{{{ e^( -.0009t ) = .9 }}}
Take the natural log of both sides
{{{ -.0009t = ln(.9) }}}
{{{ -.0009t = -.10536 }}}
{{{ t = 117.07 }}}
In 117.07 hrs, 10% have heard rumor
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Follow the same steps:
{{{ .9 = 1 - e^( -.0009t ) }}}
{{{ e^( -.0009t ) = .1 }}}
{{{ -.0009t = ln(.1) }}}
{{{ -.0009t = -2.3026 }}}
{{{ t = 2558.43 }}}
In 2558.43 hrs, 90% have heard rumor
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