How to find the initial and the future population based on today's data?
Clash Royale CLAN TAG#URR8PPP
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I need help for the part B of the following questions. Here is the question and my work:
A certain species of bird was introduced in a certain county $25$ years ago. Biologists observe that the population doubles every $10$ years, and now the population is $27,000$.
$(A)$ - What was the initial size of the bird population? (Round your answer to the nearest whole number.)
$n text(initial) = dfrac27,0002^(25/10)implies [n] text(initial) = 4773$ ----- correct.
$(B)$ - Estimate the bird population $8$ years from now. (Round your answer to the nearest whole number.)
$n text(8 years later) = 4773times 2^(8/10)implies [n] text(8 years later) = 8310$ ----- wrong.
algebra-precalculus
edited Jul 23 '15 at 3:31
L.G.
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
add a comment |Â
up vote
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down vote
favorite
I need help for the part B of the following questions. Here is the question and my work:
A certain species of bird was introduced in a certain county $25$ years ago. Biologists observe that the population doubles every $10$ years, and now the population is $27,000$.
$(A)$ - What was the initial size of the bird population? (Round your answer to the nearest whole number.)
$n text(initial) = dfrac27,0002^(25/10)implies [n] text(initial) = 4773$ ----- correct.
$(B)$ - Estimate the bird population $8$ years from now. (Round your answer to the nearest whole number.)
$n text(8 years later) = 4773times 2^(8/10)implies [n] text(8 years later) = 8310$ ----- wrong.
algebra-precalculus
edited Jul 23 '15 at 3:31
L.G.
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
1
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need help for the part B of the following questions. Here is the question and my work:
A certain species of bird was introduced in a certain county $25$ years ago. Biologists observe that the population doubles every $10$ years, and now the population is $27,000$.
$(A)$ - What was the initial size of the bird population? (Round your answer to the nearest whole number.)
$n text(initial) = dfrac27,0002^(25/10)implies [n] text(initial) = 4773$ ----- correct.
$(B)$ - Estimate the bird population $8$ years from now. (Round your answer to the nearest whole number.)
$n text(8 years later) = 4773times 2^(8/10)implies [n] text(8 years later) = 8310$ ----- wrong.
algebra-precalculus
edited Jul 23 '15 at 3:31
L.G.
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
I need help for the part B of the following questions. Here is the question and my work:
A certain species of bird was introduced in a certain county $25$ years ago. Biologists observe that the population doubles every $10$ years, and now the population is $27,000$.
$(A)$ - What was the initial size of the bird population? (Round your answer to the nearest whole number.)
$n text(initial) = dfrac27,0002^(25/10)implies [n] text(initial) = 4773$ ----- correct.
$(B)$ - Estimate the bird population $8$ years from now. (Round your answer to the nearest whole number.)
$n text(8 years later) = 4773times 2^(8/10)implies [n] text(8 years later) = 8310$ ----- wrong.
algebra-precalculus
algebra-precalculus
edited Jul 23 '15 at 3:31
L.G.
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
edited Jul 23 '15 at 3:31
L.G.
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
edited Jul 23 '15 at 3:31
L.G.
1,8921038
edited Jul 23 '15 at 3:31
L.G.
1,8921038
edited Jul 23 '15 at 3:31
L.G.
1,8921038
1,8921038
asked Jul 19 '15 at 22:55
TheNewGuy
283418
asked Jul 19 '15 at 22:55
TheNewGuy
283418
asked Jul 19 '15 at 22:55
TheNewGuy
283418
283418
1
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |Â
1
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59
1
1
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |Â
1 Answer
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Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac52=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac810=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac52=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac810=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
add a comment |Â
up vote
0
down vote
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac52=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac810=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac52=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac810=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac52=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac810=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
answered Jul 23 '15 at 3:44
Vinod Kumar Punia
2,639831
2,639831
add a comment |Â
add a comment |Â
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1
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^frac810 $
– callculus
Jul 19 '15 at 23:14
Would it be 47010?
– TheNewGuy
Jul 19 '15 at 23:18
Yes. That is what I got.
– callculus
Jul 19 '15 at 23:20
If you use the initial value of the population, you would substitute $25 + 8 = 33$ for the time since $25$ years have already elapsed. You would calculate $n(33) = 4773 cdot 2^frac3310$ to find the population eight years from now.
– N. F. Taussig
Jul 20 '15 at 9:59