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3 years ago

a country's population in 1990 was 242 million, in 1999 it was 244 million. estimate the population in 2013

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer:

247.12 million

Step-by-step explanation:

Given:

In 1990, population=242 million

In 1999, population=244 million

estimated population in 2013=?

Estimated growth is calculated by:

P(t)= Po*e^(kt)

where Po is original population

k is growth rate

t is the time in year

Now putting the values of Po=242, P(t)=244, t=9 (t=1999-1990=9) we have:

244=242e^(9k)

(244/242)=e^(9k)

ln(244/242)=9k

k=1/9ln(244/242)

k=0.00091

Now Putting k=0.00091, Po=242, and t=23 (t=2013-1990=23) we have:

P(t)=242e^(23*0.00091‬)

P(t)=247.12 million

the population in 2013 estimated to be 247.12 million!

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2 years ago

Help with my algebra homework.

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Q1. The answer is $\frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$
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Now, factorise the numerators and denominators:
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x² - 2x - 8 = x² + 2x - 4x - 2*4 = x(x+2) - 4(x+2) = (x + 2)(x - 4)
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$\frac{(x+4)(x-4)}{(x+2)(x+3)} * \frac{(x+2)(x-4)}{(x+1)(x+4)}= \frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$

Q2. The answer is $\frac{7(a-7)}{(a-8)(a+8)}$
Since a² - b² = (a-b)(a+b), then a²- 64 = a² - 8² = (a-8)(a+8).
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Answer:

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Step-by-step explanation:

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