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elena55 [62]
3 years ago
5

8, 15, 20, 23, 29, 25, 30, 24, 28, 35, 32, 28, 27, 27, 25, 28, 29, 14, 22, 18 what is the outlier?

Mathematics
1 answer:
Dominik [7]3 years ago
5 0

Answer:

17

Step-by-step explanation:

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k0ka [10]

The formula is : A=6a^2

So lets say its a cube with edge lengths of 4

We first find out the area of each side by multiplying 4*4

They you multiple the answer of that, 16, by 6, for each of the sides.

A = 96 Units ^2

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Determine whether the triangles are congruent by AA., SSS, SAS, or not similar.
klemol [59]

Step-by-step explanation:

they are similar by AA .

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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
What are the steps you would take to find the area of
kodGreya [7K]

Answer: Identify the shapes you will need to determine the area of the figure.

Calculate and add the areas of the unshaded triangle and two circles.

Step-by-step explanation:

3 0
4 years ago
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Reil [10]

Answer:

C. -9

Step-by-step explanation:

The pattern is clearly subtracting 4 each time (or adding -4).

-5 - 4 = -9

6 0
3 years ago
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