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3 years ago

5

Alan deposited $300 in an account that pays 6% interest compounded continuously. Approximately how long will it take for Alan’s

money to triple?
(Use formula A=Pe^rt where A is the accumulation amount, P is the initial amount, r is the annual rate of interest, and t is the elapsed time in years.)

Show your work for credit

1 answer:

Lelechka [254]3 years ago

3 0

9514 1404 393

Answer:

18.3 years

Step-by-step explanation:

You want ...

A/P = 3 = e^(rt) . . . for r = 0.06

Taking the natural log, this gives ...

ln(3) = 0.06t

t = ln(3)/0.06 ≈ 18.31

It will take about 18.3 years for the value to triple.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$P(U |D ) = 0.198$

b

$P(O\ n \ B) = 0.188$

c

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

=>

Also

$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

=>

So

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

So

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

So

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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3 years ago