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vaieri [72.5K]
3 years ago
13

What is the average rate of change for this exponential function for the interval from x = 1 to x = 3?

Mathematics
1 answer:
Crank3 years ago
5 0
The answer would be d
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At the beginning of the month Leslie's parents put $35.00 on her account. If lunch costs $2.55 per day, how much money would be
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$73.25

Step-by-step explanation:

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(-7/8)(-3/4)<br><br>A:6/7<br><br>B:-21/32<br><br>C:21/32
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The city has an average of 13 days of rainfall for April.
zhenek [66]

Using the Poisson distribution, we have that:

  • There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
  • There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
  • There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

For this problem, the mean is given as follows:

\mu = 13

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

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

In which:

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

P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0

P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003

P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0
2 years ago
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