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2 years ago

Find the surface area of the right trianglular prism

Mathematics

2 answers:

Vlad1618 [11]2 years ago

8 0

I can help but I need to see the problem or picture

VARVARA [1.3K]2 years ago

5 0

There is no picture to solve

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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:

$P(X$

8 0

3 years ago

50 points please help me

skelet666 [1.2K]

Answer:

Step-by-step explanation:

The slope of the line is 0 as it is parallel to x-axis.

Slope = (y2 - y1)/(x2 - x1)

= (2-2)/(4-(-2))

= 0

8 0

2 years ago

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Given that the points (-3, 2) and (1, 2) are vertices of a rectangle, what two sets of coordinates could form the other two vert

Lesechka [4]

Answer:

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explanation:

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3 years ago

the sphere at the right fits snugly inside a cude with 14in edges. what is the approximate volume of the space between the spher

ella [17]

Answer:Find the vol of the sphere:

radius of the sphere = 3 in

V = %284%2F3%29%2Api%2A3%5E3

V = 97.858

Find the vol between the sphere and cube:

216 - 97.858 = 118.142 cu/in

Step-by-step explanation:

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2 years ago

3(2^x) = 6^2x <br>How do you solve this?

iragen [17]

$3(2^x)=6^{2x}$
$3(2^x)=(3\times2)^{2x}$
$3(2^x)=3^{2x}2^{2x}$
$\ln(3(2^x))=\ln(3^{2x}2^{2x})$
$\ln3+\ln2^x=\ln3^{2x}+\ln2^{2x}$
$\ln3+x\ln2=2x\ln3+2x\ln2$
$\ln3=2x\ln3+2x\ln2-x\ln2$
$\ln3=x(2\ln3+\ln2)$
$x=\dfrac{\ln3}{2\ln3+\ln2}$

7 0

2 years ago