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snow_tiger [21]
3 years ago
12

A function w satisfies w(1) = 5 and w(4) = 1. A

Mathematics
1 answer:
kari74 [83]3 years ago
4 0

Answer: A) 1

  • Since z(3) = 4, w(z(3)) = w(4)
  • Since w(4) = 1, w(z(3)) = w(4) = 1
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t the Fidelity Credit Union, a mean of 5.8 customers arrive hourly at the drive-through window. What is the probability that, in
zhuklara [117]

Answer:

0.5217

Step-by-step explanation:

Given that:

mean \mu = 5.8

Using Poisson Probability formula:

P(X=x) = \frac{   (e^{- \mu} * {\mu^x) }}{x!}

P(X>5.8)=1-P(X \leq5)

= 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)]

= 1 -[\frac{(e^{-5.8}*5.8^0)    }{ 0!   } + \frac{(e^{-5.8}*5.8^1)    }{ 1!   } +\frac{(e^{-5.8}*5.8^2)    }{ 2!   } +\frac{(e^{-5.8}*5.8^3)    }{ 3!   } +\frac{(e^{-5.8}*5.8^4)    }{ 4!   } +\frac{(e^{-5.8}*5.8^5)    }{ 5!   } ]

=1-(0.003+0.0176+0.0509+0.0985+0.1428+0.1656)

=1-0.4783

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7 0
4 years ago
the poster you are giving your friend is in a cardboard tube with a 2 in diameter and 26 in length. will a rectangular piece of
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3 years ago
Since 1851, exactly 119 hurricanes have hit Florida (this includes the years 1851 and 2019),—only direct hits by hurricanes are
alukav5142 [94]

Answer:

There is a 0.58% probability of Florida being struck by four or more hurricanes in the same year.

Step-by-step explanation:

Since we have only the mean during the interval, we can solve this problem using the Poisson probability distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

Poisson probability distribution

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

In this problem, we have that:

Since 1851, exactly 119 hurricanes have hit Florida (this includes the years 1851 and 2019). Counting 1851 and 2019, there are 169 years in this interval. This means that \mu = \frac{119}{169} = 0.704

If the probability of hurricane strikes has remained the same since 1851, what is the probability of Florida being struck by four or more hurricanes in the same year?

This is P(X \geq 4).

Either Florida is struck by less than four hurricanes in a given year, or it is struck by 4 or more. The sum of these probabilities is decimal 1.

P(X < 4) + P(X \geq 4) = 1

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.704}*(0.704)^{0}}{(0)!} = 0.4946

P(X = 1) = \frac{e^{-0.704}*(0.704)^{1}}{(1)!} = 0.3482

P(X = 2) = \frac{e^{-0.704}*(0.704)^{2}}{(2)!} = 0.1226

P(X = 3) = \frac{e^{-0.704}*(0.704)^{3}}{(3)!} = 0.0288

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.4946 + 0.3482 + 0.1226 + 0.0288 = 0.9942

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9942 = 0.0058

There is a 0.58% probability of Florida being struck by four or more hurricanes in the same year.

7 0
3 years ago
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