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Andrei [34K]
4 years ago
15

A house purchased 5 years ago for 100,000 was sold for 161,051. Assuming exponential growth , approximate the annual growth rate

, to the nearest percent .
Mathematics
2 answers:
charle [14.2K]4 years ago
7 0

Answer:

     ln A - ln P

r = ----------------

             t

Step-by-step explanation:

The exponential growth function is A = Pe^(rt), where r is the annual growth rate as a decimal fraction and t is the number of years.

Taking the natural log of both sides, we get

ln A = ln P + rt, or

ln A - ln P = rt

Since t = 5 yr,

     ln A - ln P

r = ----------------

             t

and in this case,

     ln A - ln P         11.989-9.21034

r = ----------------  =  ------------------------ = 0.56

             t                           5

The annual growth rate, to the nearest percent, is 56%

Pepsi [2]4 years ago
3 0

Answer:

?

Step-by-step explanation:

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10²

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A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience
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Answer:

a) P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

b) P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

c) P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

X \sim Binom(n=9, p=0.99)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

We want to find this probability:

P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

Part b

P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)

P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

Part c

P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

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