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3 years ago

if a sprinkler waters 1/14 of a lawn in 1/2 an hour, how much time it will take to water the entire lawn

Mathematics

1 answer:

arlik [135]3 years ago

3 0

1/14 in 30 mins
30 mins x 14=420 mins
420 mins in hours =7 hours

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Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The mean time for the component failure is 2500 hours.

This means that $m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004$

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

$P(X > x) = e^{-0.0004*5000} = 0.1353$

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

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What is the mean of Variable A?

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5 is the mean of variable a I might be wrong

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Use partial fraction expansion to evaluate: LaTeX: \int\frac{x-1}{x^2+3x+2}dx ∫ x − 1 x 2 + 3 x + 2 d x a. LaTeX: -2\ln\left|x+1

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Whats the sum of 3/8 and 1/16 a 1/6 b 4/24 c 7/16 d 1/4

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3/8 + 1/16= 7/16
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