The populations P (in thousands) of Pittsburgh, Pennsylvania from 2000 through 2007 can be modeled by $P = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 t } }$ where t represents the year, with $t = 0$ corresponding to 2000.Use the model to find the numbers of cell sites in the year 2009.

A)2,326,853.00
B)2,327,853.00
C)2,329,853.00
D)2,328,853.00
E)2,330,853.00

Question

The populations P (in thousands) of Pittsburgh, Pennsylvania from 2000 through 2007 can be modeled by

P = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 t } }

where t represents the year, with

t = 0

corresponding to 2000.Use the model to find the numbers of cell sites in the year 2009.

A)2,326,853.00
B)2,327,853.00
C)2,329,853.00
D)2,328,853.00
E)2,330,853.00

Quizplus · Accepted Answer

The model is given by $P = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 t } }$, where t represents the year, with $t = 0$ corresponding to 2000. To find the number of cell sites in the year 2009, we need to substitute $t = 9$ into the model.$P = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 (9) } }$$P = \frac { 2632 } { 1 + 0.083 e ^ { 0.4500 } }$$P = \frac { 2632 } { 1 + 0.083 (1.5683) }$$P = \frac { 2632 } { 1 + 0.1303 }$$P = \frac { 2632 } { 1.1303 }$$P = 2,328,853.00$Therefore, the number of cell sites in the year 2009 is 2,328,853.00.

The Populations P (In Thousands) of Pittsburgh, Pennsylvania from 2000 P=26321+0.083e0.0500tP = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 t } }P=1+0.083e0.0500t2632​

The Populations P (In Thousands) of Pittsburgh, Pennsylvania from 2000 $P = \frac { 2632 } { 1 + 0.083 e ^ { 0.0500 t } }$