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ivanzaharov [21]
3 years ago
5

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

me in years.
1. By what percentage does the population increase by each year?


2. Which function is correctly rewritten to reveal the monthly growth rate of the towns population?

a P(t)= 10,000(0.87449)^t

b P(t)= 10,000(0.87449)^12t

c P(t)= 10,000(0.87449)^1/t

d P(t)= 10,000(0.87449)^t+12

Please Help
Mathematics
1 answer:
DiKsa [7]3 years ago
8 0

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

P(t)=10000(0.2)^t

1) P(0) is the population initially.

P(1) is the population after a year.

\frac{P(1)}{P(0)} represents the population increase factor.

So let's evaluate that fraction:

\frac{P(1)}{P(0)}

\frac{10000(0.2)^1}{10000(0.2)^0}

\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

P(t)=10000(0.2)^{t}

We want t to represent months.

A full year is 12 months, in a full year we have that P(1)=10000(0.2)^1=10000(0.2)=2000

So we want a new P such that P(12)=2000 since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) P(12)=10000(0.87449)^{12}=2000 \text{approximately}

b) P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}

c) P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}

d) P(12)=10000(0.87449)^{12+12}=400 \text{approximately}

So a is the function we want.

Also another way to look at this:

P(t)=10000(.2)^t where t is in years.

P(t)=10000(.2^\frac{1}{12})^t where t is in months.

And .2^\frac{1}{12}=0.874485 \text{approximately}

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