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2 years ago

13

46+ 2/7w w= -1/2 You don't need to show your work

1 answer:

Tomtit [17]2 years ago

3 0

46.1428 or -46(3/14)

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Plz help! What is the surface area of the figure?

dexar [7]

Answer:

483 ft²

Step-by-step explanation:

2(7x7)= 98

3(5x12)=180

2(12x5)=120

2(5x5)= 50

1(7x5) = <u> 35 </u>

483

6 0

3 years ago

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

Can someone please help me with this

katovenus [111]

He had 24 pounds of flour in his supply

3 0

3 years ago

Read 2 more answers

∠A and ∠B are supplementary. m∠A = (6x + 5), m∠B = (2x + 15). Find the measure of each angle.

Anastaziya [24]

Answer:

125 and 55

Step-by-step explanation:

180=8x+20

160=8x

x=20

6(20)+5 2(20)+15=

120+5=125 degrees 40+15=55 degrees

4 0

3 years ago

15 liters decreased by 60%

puteri [66]

15 divided by 60= 3.50

8 0

3 years ago