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konstantin123 [22]
3 years ago
5

Two rectangular properties share a common side. Lot A is 33 feet wide and 42 feet long.

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:

The width of lot B is 11 feet, so option A is correct.

Step-by-step explanation:

Given:

  • Two rectangular properties share a common side.  
  • Lot A is 33 feet wide and 42 feet long.  
  • The combined area of the lots = 1,848 square feet.  

To find:

How many feet wide is Lot B?  

Solution:

we know that, area of a rectangle is length x breadth

Then area of lot A = 33 x 42 = 1386 square feet.

And area of lot B = width x 42

Now, we are given that, total area = 1848  

area of lot A + area of lot B = 1848  

1386 + width x 42 = 1848  

width x 42 = 1848 – 1386  

width x 42 = 462  

width =\frac{42}{462}

width = 11

Hence, the width of lot B is 11 feet, so option A is correct.

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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
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Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

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Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

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We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

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Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

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Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

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