A local pond can only hold up to 43 geese. Eight geese are introduced into the pond. Assume that the rate of growth of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 43 - \mathrm { P } ) \mathrm { P }$ where t is time in weeks. How long will it take for the goose population to be 24?
A) 35.07 weeks
B) 58.74 weeks
C) 13.84 weeks
D) 26.50 weeks

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The Answer of A local pond can only hold up to 43 geese. Eight geese are introduced into the pond. Assume that the rate of growth of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 43 - \mathrm { P } ) \mathrm { P }$ where t is time in weeks. How long will it take for the goose population to be 24?
A) 35.07 weeks
B) 58.74 weeks
C) 13.84 weeks
D) 26.50 weeks

A Local Pond Can Only Hold Up to 43 Geese dPdt=(0.0015)(43−P)P\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 43 - \mathrm { P } ) \mathrm { P }dtdP​=(0.0015)(43−P)P

A Local Pond Can Only Hold Up to 43 Geese $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 43 - \mathrm { P } ) \mathrm { P }$