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3 years ago

8

What is the slope of the line x = 4?

2 answers:

Fofino [41]3 years ago

5 0

Answer:

Undefined

Step-by-step explanation:

If the line is strait up like x = 4 that means it is undefined.

Leviafan [203]3 years ago

3 0

Answer:

slope is undefined

Step-by-step explanation:

x = 4 is the equation of a vertical line parallel to the y- axis.

The slope of a vertical line is undefined

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

Ivan

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"

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

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

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

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

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

$P(X$

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:

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

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

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

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

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?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

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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3 years ago

Sofia has four times as many dollars as her friend jessie has. together the two have $35. how much money do they each have?

garik1379 [7]

4J = s
J + s = J + 4j = 5j = 35

J = 7
s = 28

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3 years ago

The dimensions of a 4-in. square are multiplied by 3. How is the area affected? {Show work!}

Strike441 [17]

If dimensions are multiplied by 3 then area is multiplied by 3^2

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3 years ago

What is 9*9+5\6+6+6/7-7-8+7*5+6*4*4+6

Lesechka [4]

203.03968253968254
that is the answer

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3 years ago

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Find the inverse of the function.<br> f-^1(x)= (x – 5)^3 + 1

Cloud [144]

Answer:

Step-by-step explanation:

To find the inverse of function y=f(x), swap variables and then solve for the original variable.

$y=(x-5)^3+1\\ \\ x=(y-5)^3+1\\ \\ x-1=(y-5)^3\\ \\ \sqrt[3]{x-1}=y-5\\ \\ y=\sqrt[3]{x-1}+5\\ \\ f^{-1}(x)=\sqrt[3]{x-1}+5$

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3 years ago