How to find the initial and the future population based on today's data?
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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.)
$n text{(initial)} = dfrac{27,000}{2^{(25/10)}}implies [n] text{(initial)} = 4773$ ----- 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.
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
$endgroup$
add a comment |
$begingroup$
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.)
$n text{(initial)} = dfrac{27,000}{2^{(25/10)}}implies [n] text{(initial)} = 4773$ ----- 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.
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
$endgroup$
1
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
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Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
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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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |
$begingroup$
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.)
$n text{(initial)} = dfrac{27,000}{2^{(25/10)}}implies [n] text{(initial)} = 4773$ ----- 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.
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
$endgroup$
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.)
$n text{(initial)} = dfrac{27,000}{2^{(25/10)}}implies [n] text{(initial)} = 4773$ ----- 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.
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
edited Jul 23 '15 at 3:31
L.G.
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
edited Jul 23 '15 at 3:31
L.G.
2,4741445
edited Jul 23 '15 at 3:31
L.G.
2,4741445
edited Jul 23 '15 at 3:31
L.G.
2,4741445
2,4741445
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
asked Jul 19 '15 at 22:55
TheNewGuyTheNewGuy
2831520
2831520
1
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
$begingroup$
Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
$begingroup$
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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |
1
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
$begingroup$
Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
$begingroup$
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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59
1
1
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
$begingroup$
Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
$begingroup$
Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
$begingroup$
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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59
$begingroup$
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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59
add a comment |
1 Answer
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$begingroup$
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac{5}{2}=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac{8}{10}=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac{5}{2}=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac{8}{10}=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
$endgroup$
add a comment |
$begingroup$
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac{5}{2}=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac{8}{10}=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
$endgroup$
add a comment |
$begingroup$
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac{5}{2}=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac{8}{10}=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
$endgroup$
Population is now 27000.
Before 10 yrs,it was 27000/2,
before 20 yrs it was $27000/2^2$,
before 25 yrs it was $27000/2^frac{5}{2}=27000/4sqrt2=4772$
Population after 8 yrs from now=$27000times2^frac{8}{10}=47009$
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
answered Jul 23 '15 at 3:44
Vinod Kumar PuniaVinod Kumar Punia
2,697938
2,697938
add a comment |
add a comment |
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1
$begingroup$
The population is now $27,000$. Therefore the population in eight years is $27,000cdot 2^{frac{8}{10} }$
$endgroup$
– callculus
Jul 19 '15 at 23:14
$begingroup$
Would it be 47010?
$endgroup$
– TheNewGuy
Jul 19 '15 at 23:18
$begingroup$
Yes. That is what I got.
$endgroup$
– callculus
Jul 19 '15 at 23:20
$begingroup$
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 $n(33) = 4773 cdot 2^{frac{33}{10}}$ to find the population eight years from now.
$endgroup$
– N. F. Taussig
Jul 20 '15 at 9:59