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Studentka2010 [4]
3 years ago
15

Timmy is building a new website. Right after he starts, he has 2 visitors. He expects his visitors to increase according to a mo

del he developed for his other websites growth, v(t)=10t+b, where b is the number of visitors right afte he starts building his website, t is the time since he stgarted building his website in weeks and v(t) is the nunber of total visitors after t weeks.
How many visitors does he get 4 weeks after starting to build his website?

How any new visitors does he get per week?
Mathematics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

He gets 42 visitors 4 weeks after starting to build his website.

He gets 10 new visitors per week.

Step-by-step explanation:

Equation for the number of visitors:

The equation for the number of visitors Timmy's new website receives after t weeks is:

v(t) = 10t + b

In which b is the number of visitors rightly after he starts.

Timmy is building a new website. Right after he starts, he has 2 visitors.

This means that b = 2, so:

v(t) = 10t + 2

How many visitors does he get 4 weeks after starting to build his website?

This is v(4). So

v(4) = 10(4) + 2 = 40 + 2 = 42

He gets 42 visitors 4 weeks after starting to build his website.

How any new visitors does he get per week?

After 0 weeks:

v(0) = 10(0) + 2 = 0 + 2 = 2

After 1 week:

v(1) = 10(1) + 2 = 10 + 2 = 12

2 weeks:

After 2 week:

v(2) = 10(2) + 2 = 20 + 2 = 22

22 - 12 = 12 - 2 = 10

He gets 10 new visitors per week.

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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.

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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}.

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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}.

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

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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.

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