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

question 36

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) \(\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 33 \mathrm { e } ^ { - 0.051 \mathrm { t } } }\)

Grasp the impact of incorrect classification of expenditures on financial statements.

Comprehend the process of recording disposals and exchanges of fixed assets.

Identify the elements of depreciation expense computation and their variations based on the method applied.

Identify common names and effects of diverse drugs including cocaine, amphetamines, PCP, and LSD.

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