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a c ountrys population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponetial growth formula. round your answer to the nearest million.

Guest Dec 1, 2014

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 #1
avatar+18827 
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a c ountrys population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponetial growth formula. round your answer to the nearest million.

$$\begin{array}{rcl}
(1) \quad p(1991) = 231 &=& p_0 * e^{\lambda*1991}\\
(2) \quad p(1999) = 233 &=& p_0 * e^{\lambda*1999} \\
\hline
\end{array}\\\\
(2):(1)
\begin{array}{rcl}
\frac{233}{231} &=& \frac{ \not{p_0} * e^{\lambda*1999} } {\not{p_0} * e^{\lambda*1991} }\\\\
\frac{233}{231}& = &e^{\lambda*1999-\lambda*1991} = e^{8\lambda}\\\\
\ln{(\frac{233}{231})&=& 8\lambda} \\\\
\lambda &=& \frac{ \ln{(\frac{233}{231} )} } {8} \\\\
\textcolor[rgb]{1,0,0}{ \lambda = 0.00107759288 }
\end{array}\\$$

 

$$\\p_0 = \dfrac{231}{ e^{\lambda*1991} } = \dfrac{231}{ e^{0.00107759288*1991} } \\\\
\textcolor[rgb]{1,0,0}{p_0= 27.0295384716}$$

exponetial growth formula: $$\boxed{p(year) = 27.0295384716 * e^{ 0.00107759288 * year}}$$

$$\\p(\textcolor[rgb]{1,0,0}{2003}) = 27.0295384716 * e^{ 0.00107759288 * \textcolor[rgb]{1,0,0}{2003} }\\\\
p(2003)= 27.0295384716 * e^{2.15841853962}\\\\
p(2003)= 27.0295384716 * 8.65743543541\\\\
p(2003)= 234.006484167\\\\
\boxed{p(2003)\approx 234 \ Million}$$

heureka  Dec 1, 2014
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1+0 Answers

 #1
avatar+18827 
+5
Best Answer

a c ountrys population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponetial growth formula. round your answer to the nearest million.

$$\begin{array}{rcl}
(1) \quad p(1991) = 231 &=& p_0 * e^{\lambda*1991}\\
(2) \quad p(1999) = 233 &=& p_0 * e^{\lambda*1999} \\
\hline
\end{array}\\\\
(2):(1)
\begin{array}{rcl}
\frac{233}{231} &=& \frac{ \not{p_0} * e^{\lambda*1999} } {\not{p_0} * e^{\lambda*1991} }\\\\
\frac{233}{231}& = &e^{\lambda*1999-\lambda*1991} = e^{8\lambda}\\\\
\ln{(\frac{233}{231})&=& 8\lambda} \\\\
\lambda &=& \frac{ \ln{(\frac{233}{231} )} } {8} \\\\
\textcolor[rgb]{1,0,0}{ \lambda = 0.00107759288 }
\end{array}\\$$

 

$$\\p_0 = \dfrac{231}{ e^{\lambda*1991} } = \dfrac{231}{ e^{0.00107759288*1991} } \\\\
\textcolor[rgb]{1,0,0}{p_0= 27.0295384716}$$

exponetial growth formula: $$\boxed{p(year) = 27.0295384716 * e^{ 0.00107759288 * year}}$$

$$\\p(\textcolor[rgb]{1,0,0}{2003}) = 27.0295384716 * e^{ 0.00107759288 * \textcolor[rgb]{1,0,0}{2003} }\\\\
p(2003)= 27.0295384716 * e^{2.15841853962}\\\\
p(2003)= 27.0295384716 * 8.65743543541\\\\
p(2003)= 234.006484167\\\\
\boxed{p(2003)\approx 234 \ Million}$$

heureka  Dec 1, 2014

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