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The U.S. population in 1990 was approximately 250 million, and the average growth rate for the past 30 years gives a doubling t...Asked by Angela
The U.S. population in 1990 was approximately 250 million, and the average growth
rate for the past 30 years gives a doubling time of 66 years. The above formula for the
United States then becomes
P (in millions)= 250 x 2( y-1990)/66
1.What was the approximate population of the United States in 1960?
2. What will the population of the United States be in 2025 if this
growth rate continues?
P (in millions)= 250 x 2^[( y-1990)/66]
Insert y = 1960 and y = 2025 in here.
yes i do that but I do not get a correct answer. 2025-1990=35/66=.5303
1960-1990=-30/36=-.4545
what do i do from there I know the answer for the 1960 in 182 million but I am still stuck on how they got that answer
250*2^(0.5303) = ...? (use your calculator)
And:
250 million *2^(-0.4545) = 182 million
So it is the square root -0.4545
i appreciate your help
No, it's 2 to the power -0.4545.
I think you know that:
a^n = a*a*a*... *a (n factors of a)
But here n must be a positive integer. It turns out one can "continue" the power function allowing n to take on any real value and not just a positive integer.
I know I am asking a lot of question but let say -0.4545, how exactly do you figure that out, I do not have a calucator that I can push in the numbers, I use the one on my computer, that is why I am asking.
Thank You Again For your Help
If you have the e^x (or Exp) function and the ln(x) function, then you can use that:
2^(x) = e^[x*ln(2)] Or
2^(x) = Exp[x*ln(2)]
If you have just addition multiplication and division, then you can still compute it using series expansions. I can explain that later, if necessary...
rate for the past 30 years gives a doubling time of 66 years. The above formula for the
United States then becomes
P (in millions)= 250 x 2( y-1990)/66
1.What was the approximate population of the United States in 1960?
2. What will the population of the United States be in 2025 if this
growth rate continues?
P (in millions)= 250 x 2^[( y-1990)/66]
Insert y = 1960 and y = 2025 in here.
yes i do that but I do not get a correct answer. 2025-1990=35/66=.5303
1960-1990=-30/36=-.4545
what do i do from there I know the answer for the 1960 in 182 million but I am still stuck on how they got that answer
250*2^(0.5303) = ...? (use your calculator)
And:
250 million *2^(-0.4545) = 182 million
So it is the square root -0.4545
i appreciate your help
No, it's 2 to the power -0.4545.
I think you know that:
a^n = a*a*a*... *a (n factors of a)
But here n must be a positive integer. It turns out one can "continue" the power function allowing n to take on any real value and not just a positive integer.
I know I am asking a lot of question but let say -0.4545, how exactly do you figure that out, I do not have a calucator that I can push in the numbers, I use the one on my computer, that is why I am asking.
Thank You Again For your Help
If you have the e^x (or Exp) function and the ln(x) function, then you can use that:
2^(x) = e^[x*ln(2)] Or
2^(x) = Exp[x*ln(2)]
If you have just addition multiplication and division, then you can still compute it using series expansions. I can explain that later, if necessary...
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