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timama [110]
1 year ago
11

A rectangle with a length of 2 w and width of w. An equilateral triangle with side lengths of w + 8. The perimeters of a rectang

le and an equilateral triangle are equal. What is the equation for this scenario? rectangle’s perimeter: equilateral triangle’s perimeter: scenario’s equation:
Mathematics
2 answers:
Mkey [24]1 year ago
8 0

Answer:

rectangles perimeter: 6w

equilateral triangles perimeter: 3w+24

scenario’s equation: 6w=3w+24

Step-by-step explanation:

w=8

rectangles perimeter= 48

EQUTRI perimeter= 48

inn [45]1 year ago
7 0

Answer:

1st part: 3rd option

2nd part: 4th option

3rd part: 4th option

Step-by-step explanation:

Which expression represents the rectangle’s perimeter?

2(2w-6)+2(w)

Which expression represents the triangle’s perimeter?

3(w+8)

What is the perimeter of both polygons?

60 units

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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t
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Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

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

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 interval.

A bank gets an average of 12 customers per hour.

This means that \mu = 12

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

P(X < 10) + P(X \geq 10) = 1

We want P(X \geq 10)

Then

P(X \geq 10) = 1 - P(X < 10)

In which

P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

So

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

P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0

P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001

P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004

P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018

P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053

P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127

P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255

P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437

P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655

P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874

P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0 + 0.0001 + 0.0004 + 0.0018 + 0.0053 + 0.0127 + 0.0255 + 0.0437 + 0.0655 + 0.0874 = 0.2424

Then

P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576

75.76% probability that there will be 10 or more customers at this bank in one hour.

3 0
2 years ago
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