In the year 2000, the population of a city was 600,000 citizens. The population increases at a rate of 1.8% per year.
1 answer:
Answer:
a) 600,000 (1.018)ᵗ, the graph is below:
* "t" is the years. 2000 is the first year, so "t" would be 1. 2012 is the 12th year so "t" would be 12.
b) As the years go by, the population increases. The years (X), and population (Y) both increase together. The graph shows that when the years increase the population increases.
c) In 2012, which would be 12 years later, I predict the population would be around 743,232 people.
Step-by-step explanation:
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Answer:
Step-by-step explanation:
we know that
The compound interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
substitute in the formula above
Answer:
2
Step-by-step explanation:
we start simplifying by removing the parenthesis
Multiply the exponents inside the the parenthesis
3^4 * 2^4
Now we apply exponential property
a^m * a^n = a^ (m+n)
3^4 * 3^-3 = 3^ (4-3) = 3^1
3 or 3^1 are same
3^1 at the top and bottom so we cancel it out
\frac{2^4}{2^3}
we apply log property . a^m / a^n = a^m-n
Now subtract the exponents
2^(4-3) = 2^1 = 2