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2 years ago

Please help me with this

Mathematics

1 answer:

charle [14.2K]2 years ago

8 0

No, you need to study. Listen to your teach and start sucking my dix you woore

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The following table shows the estimated populations and annual growth rates for four countries in the year 2000. Find the

gladu [14]

Answer:

Part 1) Australia $19,751,012\ people$

Part 2) China $1,319,645,764\ people$

Part 3) Mexico $109,712,539\ people$

Part 4) Zaire $60,534,681\ people$

Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

$P(t)=a(1+r)^t$

where

P(t) is the population

t is the number of years since year 2000

a is he initial value

r is the rate of change

Part 1) Australia

we have

$a=19,169,000\\r=0.6\%=0.6\100=0.006$

substitute

$P(t)=19,169,000(1+0.006)^t$

$P(t)=19,169,000(1.006)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=19,169,000(1.006)^5=19,751,012\ people$

Part 2) China

we have

$a=1,261,832,000\\r=0.9\%=0.9\100=0.009$

substitute

$P(t)=1,261,832,000(1+0.009)^t$

$P(t)=1,261,832,000(1.009)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=1,261,832,000(1.009)^5=1,319,645,764\ people$

Part 3) Mexico

we have

$a=100,350,000\\r=1.8\%=1.8\100=0.018$

substitute

$P(t)=100,350,000(1+0.018)^t$

$P(t)=100,350,000(1.018)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=100,350,000(1.018)^5=109,712,539\ people$

Part 4) Zaire

we have

$a=51,965,000\\r=3.1\%=3.1\100=0.031$

substitute

$P(t)=51,965,000(1+0.031)^t$

$P(t)=51,965,000(1.031)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=51,965,000(1.031)^5=60,534,681\ people$

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