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frez [133]
4 years ago
15

alex needs to varnish the top and the bottom of a dozen rectangular wooden planks.The planks are 8 feet long and 3 feet wide.Eac

h pint of varnish covers about 125 square feet and costs $3.50. A:What is the total area that alex needs to varnish? B:How much will it cost alex to varnish all the wooden planks?
Mathematics
1 answer:
Marta_Voda [28]4 years ago
3 0
Given:
12 rectangular planks
2 sides per plank
Length = 8 ft
Width = 3 ft

Area of a rectangle = Length * Width.
A = 8ft * 3ft
A = 24 ft²

24 ft² x 2 sides of a plank = 48 ft²
48 ft²/plank * 12 planks = 576 ft² is the total area Alex needs to varnish.

576 ft² ÷ 125 ft² = 4.608 pint or round to 5 pints

5 pints x $3.50/pint = $17.50 Total cost that Alex need to spend to varnish all the wooden planks.



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The city has an average of 13 days of rainfall for April.
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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

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2 years ago
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