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Montano1993 [528]
4 years ago
8

The population of a county is growing at a rate of 9% per year, compounded continuously. How many years will it take for the pop

ulation to quadruple according to the exponential growth function?
Mathematics
1 answer:
torisob [31]4 years ago
8 0

Answer:

t=15.4\ years

Step-by-step explanation:

The  exponential growth function compounded continuously is equal to

A=P(e)^{rt}  

where  

A is the final population  

P is the initial population  

r is the rate of growth in decimal  

t is Number of years

e is the mathematical constant number

we have  

A=4x\\P=x\\ r=9\%=9/100=0.09  

substitute in the function above

4x=x(e)^{0.09t}    

simplify

4=(e)^{0.09t}

Take natural log of both sides

ln(4)=ln[(e)^{0.09t}]

ln(4)=0.09t(ln(e))

ln(e)=1

ln(4)=0.09t

t=ln(4)/0.09

t=15.4\ years

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Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t
bogdanovich [222]

Answer:

a) \sigma_{\bar x} = 1.414

b) \sigma_{\bar x} = 1.414

c) \sigma_{\bar x} = 1.414

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Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation  =10

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a) The population size is infinite;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

c)  When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }

\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }

\sigma _{\bar x} = 1.343

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The given equation is

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To solve for x, we have to subtract 8 to both sides

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Now we need to get rid of -9 from the left side, and for that we do division

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