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3 years ago

The principal $3000 is accumulated with 3% interest, compounded semiannually for 6 years.

1 answer:

7 0

The formula is
A=p (1+r/k)^kt
A accumulated amount?
P principle 3000
R interest rate 0.03
K compounded semiannually 2
T time 6 years
A=3,000×(1+0.03÷2)^(2×6)
A=3,586.85

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There are 10 women and 3 men in room A. One person is picked at random from room A and moved to room B, where there are already

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Answer:

The probability that a woman will be picked from room B is 0.34.

Step-by-step explanation:

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The lifetime of a mechanical assembly in a vibration test is exponentially distributed with a mean of 500 hours. If an assembly

skelet666 [1.2K]

Answer:

A) probability of failure in next 100 hours given that it has been tested for 500 hours without failure is 0.181

B) probability that exactly two have the metabolic defect is 0.03

Step-by-step explanation:

Part A)

Let X be a exponentially random variable with mean = μ = 500 hrs

For exponential distribution:

$p.d.f = f(x) = \lambda e^{-\lambda x}\\c.d.f = F(x) = 1 - e^{-\lambda x}\\x\geq 0$

λ = 1/μ

λ = 0.002

We have to find the probability of failure in the next 100 hours given that assembly has been tested for 500 hours without a failure.

Using memory less property of exponential distribution:

$P(X500) = P (X$

using

$F(x) = 1 - e^{-\lambda x}\\ \lambda =.002\\x=100\\F(x) = 1- e^{-(.002)(100)}\\F(x) = 1-.8187\\F(x) = 0.181$

Chances of occurrence of metabolic defect = 5%

P(C) = .05

No. of randomly selected infants = n =6

We have to find the probability that exactly two have the metabolic defect

⇒x = 2

Using binomial probability density function:

P = $P=\left[\begin{array}{ccc}n\\x\end{array}\right] p^{x} (1-p) ^{n-x}\\\\=\frac{n!}{x!(n-x)!} p^{x} (1-p) ^{n-x}\\=\frac{6!}{2!4!}(.05)^{2}(.95)^{4}\\= 0.03\\$

probability that exactly two have the metabolic defect is 0.03

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