2 between 1.3 and 1.4

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0

3 years ago

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Help please !!!!thank yo

goldfiish [28.3K]

Answer:

250

Step-by-step explanation:

There is a 200 start up cost plus 50 per month

Cost = 200 + 50m where m is the number of months

We have 1 month

Cost = 200 +50(1)

Cost = 250

6 0

3 years ago

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PLEASE HELP ME!!!!<br><br> (if you don’t know don’t answer)

Sergio039 [100]

Answer:

Here are my answers

hope this helps

Step-by-step explanation:

1. isosceles triangle

2. 3x + 15 = 90 x = 25

x + 25 = 90 x = 65

x + 55 = 90 x = 35

3. The size for DAB is 84

My reasoning is I went from one angle to another by subtracting 180 from my answers. For example angle 132 is supplementary to the missing angle next to it. I subtract 180 from 132 and get 48. That already solves the 2 equivalent angles in the triangle so I have 48+48+x = 180 because a full triangle equals 180. I get x = 84 then I do the same thing I did with angle 132. Angle 84 is supplementary to the missing angle next to it. I subtract 84 from 180 and get 96 then go diagonal from angle C (which is 96) to angle A. Those are supplementary so I subtract 96 from 180 and get 84.

3 0

3 years ago

Can someone help me please!??

EleoNora [17]

Is D (0,2) ...........

4 0

3 years ago

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If 4 apples cost 0.80, how much do 12 apples cost

Neporo4naja [7]

4 / 0.80 = 12 / x...cross multiply
4x = 0.80 * 12
4x = 9.6
x = 9.6 / 4
x = 2.4.....12 apples cost $ 2.40

3 0

3 years ago

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