Solve
-A Local Pond Can Only Hold Up to 39 $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 39 - \mathrm { P } ) \mathrm { P }$

Multiple Choice

Solve.
-A local pond can only hold up to 39 geese. Six geese are introduced into the pond. Assume that the rate of grow of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 39 - \mathrm { P } ) \mathrm { P }$ where $t$ is time in weeks. Find a formula for the goose population in terms of $\mathrm { t }$ .

A) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 5.5 \mathrm { e } ^ { - 0.051 \mathrm { t } } }$
B) $P ( t ) = \frac { 39 } { 1 + 33 e ^ { - 39 t } }$
C) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 5.5 \mathrm { e } ^ { - 39 \mathrm { t } } }$
D) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 33 \mathrm { e } ^ { - 0.051 \mathrm { t } } }$