a small (fictional) country had a population of 2,254,000 people in the year 2000. If the population grows steadily at a rate of 3.5% each year, what is the expected population of this country in the year 2015 to the nearest 1000?
Since each year the population is multiplied by 1.035, after 15 years it will be (in thousands)
2254 * 1.035^15
The answer will be
1,183,000
To get that you will do this set up
Is %
— *. —
Of. 100
3.5 is the percentage
100 stays 100
If is 2,254,000, you know that because the word “of” is in front of the number
Is is well we don’t know
Know the set up will look like this
X. 3.5
————— * ——
2,254,000. 100
Cross multiply
3.5* 2,254,00= 7,889,000
100 *x= 100x
Now do 100/x=x
And now do 7,889,000/100= 78,890
Now multiply that number by how many years has went by then round the number to the nearest 1000
1,183,000 people in that country in 2015
Sorry wrong person
To calculate the expected population in the year 2015, we need to understand the concept of compound interest. In this case, the population is growing at a constant rate of 3.5% each year.
To calculate the expected population, we need to use the formula for compound interest:
A = P(1 + r)^n
Where:
A = the final amount (expected population in the year 2015)
P = the initial amount (population in the year 2000)
r = the annual growth rate (3.5% or 0.035 as a decimal)
n = the number of years (2015 - 2000 = 15)
Now let's calculate the expected population:
P = 2,254,000
r = 0.035
n = 15
A = P(1 + r)^n
A = 2,254,000(1 + 0.035)^15
A ≈ 2,254,000(1.035)^15
A ≈ 2,254,000(1.5878526407)
Now, let's calculate the approximate value of the expected population by rounding to the nearest 1000:
A ≈ 2,254,000(1.5878526407)
A ≈ 3,579,565.1232084
Rounding this value to the nearest 1000, we get:
A ≈ 3,580,000
Therefore, the expected population of the fictional country in the year 2015, to the nearest 1000, is approximately 3,580,000.