The population of a village
The population of a village was $1000$ one year ago. The population at present is $10210$. Find the population growth rate.
My attempt:
Initial Population $(P_0)=1000$
Final Population $(P_n)=10210$
Time $(T)=1$ year
Rate $(R)=?$
Now,
$$P_n=P_0left(1+frac {R}{100}right)^T$$
$$frac {1021}{100}=left(1+frac {R}{100}right)$$
But the answer given in my book is $2%$. Why do the answers not match?
algebra-precalculus
edited Nov 27 at 6:49
Robert Howard
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
add a comment |
The population of a village was $1000$ one year ago. The population at present is $10210$. Find the population growth rate.
My attempt:
Initial Population $(P_0)=1000$
Final Population $(P_n)=10210$
Time $(T)=1$ year
Rate $(R)=?$
Now,
$$P_n=P_0left(1+frac {R}{100}right)^T$$
$$frac {1021}{100}=left(1+frac {R}{100}right)$$
But the answer given in my book is $2%$. Why do the answers not match?
algebra-precalculus
edited Nov 27 at 6:49
Robert Howard
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
2
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33
add a comment |
The population of a village was $1000$ one year ago. The population at present is $10210$. Find the population growth rate.
My attempt:
Initial Population $(P_0)=1000$
Final Population $(P_n)=10210$
Time $(T)=1$ year
Rate $(R)=?$
Now,
$$P_n=P_0left(1+frac {R}{100}right)^T$$
$$frac {1021}{100}=left(1+frac {R}{100}right)$$
But the answer given in my book is $2%$. Why do the answers not match?
algebra-precalculus
edited Nov 27 at 6:49
Robert Howard
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
The population of a village was $1000$ one year ago. The population at present is $10210$. Find the population growth rate.
My attempt:
Initial Population $(P_0)=1000$
Final Population $(P_n)=10210$
Time $(T)=1$ year
Rate $(R)=?$
Now,
$$P_n=P_0left(1+frac {R}{100}right)^T$$
$$frac {1021}{100}=left(1+frac {R}{100}right)$$
But the answer given in my book is $2%$. Why do the answers not match?
algebra-precalculus
algebra-precalculus
edited Nov 27 at 6:49
Robert Howard
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
edited Nov 27 at 6:49
Robert Howard
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
edited Nov 27 at 6:49
Robert Howard
1,9161822
edited Nov 27 at 6:49
Robert Howard
1,9161822
edited Nov 27 at 6:49
Robert Howard
1,9161822
1,9161822
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
asked Nov 12 '16 at 8:08
blue_eyed_...
3,23821645
3,23821645
2
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33
add a comment |
2
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33
2
2
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33
add a comment |
1 Answer
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It looks like Claude's insight was correct; replacing $10210$ with $1020$ does yield a growth rate of $2%$.
begin{align}
P_n&=P_0left(1+frac{R}{100}right)^T \
frac{1020}{1000}&=1+frac{R}{100} \[0.8ex]
frac{2}{100}&=frac{R}{100} \[0.8ex]
R&=2
end{align}
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
It looks like Claude's insight was correct; replacing $10210$ with $1020$ does yield a growth rate of $2%$.
begin{align}
P_n&=P_0left(1+frac{R}{100}right)^T \
frac{1020}{1000}&=1+frac{R}{100} \[0.8ex]
frac{2}{100}&=frac{R}{100} \[0.8ex]
R&=2
end{align}
add a comment |
It looks like Claude's insight was correct; replacing $10210$ with $1020$ does yield a growth rate of $2%$.
begin{align}
P_n&=P_0left(1+frac{R}{100}right)^T \
frac{1020}{1000}&=1+frac{R}{100} \[0.8ex]
frac{2}{100}&=frac{R}{100} \[0.8ex]
R&=2
end{align}
add a comment |
It looks like Claude's insight was correct; replacing $10210$ with $1020$ does yield a growth rate of $2%$.
begin{align}
P_n&=P_0left(1+frac{R}{100}right)^T \
frac{1020}{1000}&=1+frac{R}{100} \[0.8ex]
frac{2}{100}&=frac{R}{100} \[0.8ex]
R&=2
end{align}
It looks like Claude's insight was correct; replacing $10210$ with $1020$ does yield a growth rate of $2%$.
begin{align}
P_n&=P_0left(1+frac{R}{100}right)^T \
frac{1020}{1000}&=1+frac{R}{100} \[0.8ex]
frac{2}{100}&=frac{R}{100} \[0.8ex]
R&=2
end{align}
answered Nov 27 at 6:27
community wiki
Robert Howard
add a comment |
add a comment |
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2
I suppose one more typo in a textbook ! $10210$ should be $1020$ to get the "answer"
– Claude Leibovici
Nov 12 '16 at 8:13
Is it how many people per month? Per year?
– Obinna Nwakwue
Jun 17 '17 at 20:33