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3 years ago

Use the distance formula to find distance between two points. Need help on this ASAP.

Mathematics

1 answer:

astra-53 [7]3 years ago

3 0

Answer:

a) 20 units

b) 2 √10 units

c) 2 √17 units

Step-by-step explanation:

The distance formula is;

D = √(y2-y1)^2 + (x2-x1)^2

a) D = √(-7-9)^2 + (-7-5)^2

D = √256 + 144

D = √400

D = 20

b) D = √(10-8)^2 + (9-2)^2

D = √(2)^2 + 6^2

D = √4 + 36)

D = √40

D = 2 √10 units

c) D = √(1 + 7)^2 + (-8+10)^2

D = √(64 + 4)

D = √68

D = 2 √17 units

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2 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:

$P(X$

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4 years ago

I need help here’s a picture can somone explain what I need to do??

BigorU [14]

The total area of the composite figure is 176 ft

<h3>How to find the area of a composite figure?</h3>

The area of a composite figure can be found as follows;

Therefore,

Total area = area of rectangle + area of triangle

area of rectangle = lw

where

l = length
w = width

Therefore,

area of a rectangle = 16 × 8

area of a rectangle = 128 ft²

area of triangle = 1 / 2 bh

where

b = base
h = height

Therefore,

area of triangle = 1 / 2 × 12 × 8

area of triangle = 48 ft²

Total area of the composite figure = 48 + 128 = 176 ft

learn more on composite figure here: brainly.com/question/12315384

#SPJ1

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2 years ago