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alexira [117]
2 years ago
8

The population of a small town is decreasing exponentially at a rate of 14.3% each year. The current population is 9,400 people.

The town's tax status will change once the population is below 6,000 people.
Create an inequality that can be used to determine after how many years, t, the town's tax status will change, and use it to answer the question below.
Will the town's tax status change within the next 3 years?
Mathematics
1 answer:
KonstantinChe [14]2 years ago
7 0

Using an exponential function, the inequality is given as follows:

9400(0.857)^t < 6000

The solution is t > 2.9, hence the tax status will change within the next 3 years.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

A(t) = A(0)(1 - r)^t

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

For this problem, the parameters are given as follows:

A(0) = 9400, r = 0.143.

The population after t years is modeled by:

A(t) = A(0)(1 - r)^t

A(t) = 9400(1 - 0.143)^t

A(t) = 9400(0.857)^t

The tax status will change when:

A(t) < 6000

Hence the inequality is:

9400(0.857)^t < 6000

Then:

(0.857)^t < \frac{6000}{9400}

\log{(0.857)^t} < \log{\left(\frac{6000}{9400}\right)}

t\log{0.857} < \log{\left(\frac{6000}{9400}\right)}

Since both logs are negative:

t > \frac{\log{\left(\frac{6000}{9400}\right)}}{\log{0.857}}

t > 2.9.

The solution is t > 2.9, hence the tax status will change within the next 3 years.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

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Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

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Step-by-step explanation:

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If we want to proof this we need to show that for any number is true.

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