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4 years ago

5

Find the greatest common factor of 270 and 360. (Give the answer in the numerical form in the top box and in exponential form by

filling in the boxes for exponents.)

1 answer:

Andreyy894 years ago

5 0

270, 360 | 2
135, 180 | 3
45, 60 | 3
15, 20 | 5
3, 4

2 * 3 * 3 * 5 = 90
The greatest common factor of 270 and 360 is 90.

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At 5:45 p.m., a jet is located 108 mi due east of a city. A second jet is located 214 mi due north of the city. To the nearest t

siniylev [52]

By connecting the distances traveled by the jet and the distance between them, we form a right triangle with hypotenuse equal to the unknown distance.
Using the Pythagorean,
h² = a² + b²
Substituting,
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PLEASE PLEASE HELP ME A pinecone drops from a tree branch that is 36 feet above the ground. The function h = -16t2 + 36 is used.

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Answer:

T = 1.5s

Step-by-step explanation:

Hello,

To find the time the pincone hits the ground, we need to use the equation given.

Note that h = 0 when the pinecones hits the ground.

This question relates to motion under gravity.

h = -16t² + 36

0 = -16t² + 36

Make t² the subject of formula

16t² = 36

t² = 36 / 16

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Take the square root of both sides

t = √(2.25)

t = 1.5s

The time it takes the pinecone to hit the ground is 1.5s.

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3 years ago

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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:

$P(X$

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