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# In 2012, the growth of rate of the south african population was -0,412% Determine the population of South Africa in 2011 if the population w

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In 2012, the growth of rate of the south african population was -0,412% Determine the population of South Africa in 2011 if the population was 48, 810, 427 in 2012

Jul 29, 2015

#1
+27374
+10

Rate of growth is dN/dt = -0.412/100*N which can be solved to get  Nt = N12*e-0.412/100*t where N12 = 48810427.  (I've assumed the -0.412% means -0.412% per year)

So, to find N11 (the number  in 2011) we set t = -1 (i.e. it's 1 year earlier) to get:

$${\mathtt{N11}} = {\mathtt{48\,810\,427}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{0.412}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right)\right)} \Rightarrow {\mathtt{N11}} = {\mathtt{49\,011\,940.792\: \!604\: \!870\: \!237\: \!79}}$$

or N11 ≈ 49011941

.

Jul 29, 2015

#1
+27374
+10

Rate of growth is dN/dt = -0.412/100*N which can be solved to get  Nt = N12*e-0.412/100*t where N12 = 48810427.  (I've assumed the -0.412% means -0.412% per year)

So, to find N11 (the number  in 2011) we set t = -1 (i.e. it's 1 year earlier) to get:

$${\mathtt{N11}} = {\mathtt{48\,810\,427}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{0.412}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right)\right)} \Rightarrow {\mathtt{N11}} = {\mathtt{49\,011\,940.792\: \!604\: \!870\: \!237\: \!79}}$$

or N11 ≈ 49011941

.

Alan Jul 29, 2015
#2
+5

Thank you so much Alan ^_^

Jul 29, 2015
#3
+95356
+5

Thanks Alan

Hi anon, you probably already realize this but the growth rate is negative so the population is declining - not actually growing.

Jul 29, 2015