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2 years ago

Answer the question pls xx

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

3 0

Answer:

3 lines of symmetry

Step-by-step explanation:

the lines go through the vertices of the triangle

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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

The area of a sector of an circle of a radius 5 cm, formed by an arc of length is 3.5cm. what is the answer

Natali5045456 [20]

Answer:

8.75cm²

Step-by-step explanation:

First we need to find the angle subtended by the arc

L = theta/360 * 2pir

3.5 = theta/360 * 2*3.14(5)

3.5 = theta/360 * 31.4

theta/360 = 3.5/31.4

theta/360 = 0.11146

theta = 0.11146 * 360

theta = 40.13 degrees

Area of the sector = theta/360 * pir^2

Area of the sector = 40.13/360 * 3.14*25

Area of the sector = 40.13/360 * 78.5

Area of the sector = 8.75cm

Hence the area of the sector is 8.75cm²

5 0

2 years ago

If z=15, what would the expression z2 – 2z + 15 be equal to?

svet-max [94.6K]

Answer:

Step-by-step explanation:

z2 - 2z + 15

(15)2 - 2(15) + 15

30 - 30 + 15

0 + 15

5 0

2 years ago

Read 2 more answers

Dimitri earns $4.50 for each box of fruit picked

Gemiola [76]

A) $90
b) 33.34 (round to 34 if you need a solid number)
c) $13.50 an hour

5 0

2 years ago

Complete the equation of the line whose slope is 5 and y intercept is (0,4)

irina1246 [14]

y = 5x + 4 is the equation of the line whose slope is 5 and y intercept is (0,4)

Solution:

Given that, we have to write the equation of the line whose slope is 5 and y intercept is (0,4)

The equation of line in slope intercept form is given as:

y = mx + c ---- eqn 1

Where, "m" is the slope of line and "c" is the y - intercept

Given that, slope = m = 5

y intercept is (0, 4)

So, c = 4

Substitute c = 4 and m = 5 in eqn 1

y = 5x + 4

Thus the equation of line is found

7 0

3 years ago