Question 946222
When {{{t=0}}},{{{Q=1900}}}
Find {{{t}}} when, {{{Q=(1/2)1900}}}
{{{(1/2)1900=1900e^(-0.025t[h])}}}
{{{ln(1/2)=-0.025t[h]}}}
{{{t[h]=-(ln(1/2)/0.025)}}}
approximately,
{{{t[h]=27.73}}}