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3 years ago

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A bank account has a beginning balance of $560.00. After 6 months, the balance in the account has increased to $572.60. What int

erest rate is being earned annually on this account?

1 answer:

nikdorinn [45]3 years ago

8 0

The question to the question is....................

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Kisachek [45]

10 squared minus 8 squared.

=36

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3 years ago

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Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ=0

Pepsi [2]

Answer:

The probability that a repair time exceeds 8 hours is $P(X>8)\approx 0.000746586$

The conditional probability that a repair takes at least 13 hours, given that it takes more than 9 hours is $P(X\geq 13 | X>9)\approx 0.0273237376$

Step-by-step explanation:

A continuous random variable X is said to have an exponential distribution with parameter $\lambda > 0$ shown as $X \sim Exponential(\lambda)$ , if its probability density function is given by

$f_X(x)=\begin{cases}\lambda e^{-\lambda x} & x > 0\\ 0 &\text{otherwise}\end{cases}$

Let X denote the time require to repair a machine.

(a) The probability that a repair time exceeds 8 hours;

$P(X>8)=1-P(X\leq 8)\\P(X>8)=1-\int\limits^8_0 {0.9e^{-0.9x}} \, dx \\P(X>8)=1-0.999253\\P(X>8)\approx 0.000746586$

(b) The conditional probability that a repair takes at least 13 hours, given that it takes more than 9 hours;

We want to find $P(X\geq 13 | X>9)$ .

$P(X\geq 13 | X>9)=\frac{P(X\geq 13)}{P(X>9)} \\\\P(X\geq 13 | X>9)=\frac{\int\limits^{\infty}_{13} {0.9e^{-0.9x}} \, dx}{\int\limits^{\infty}_{9} {0.9e^{-0.9x}} \, dx}\\ \\P(X\geq 13 | X>9)=\frac{8.29382\times10^{-6}}{0.000303539} \\\\P(X\geq 13 | X>9)\approx 0.0273237376$

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3 years ago