X-7/4 =-2 I need helpp

What is the sum?

Nezavi [6.7K]

Answer:

Its B on e2020

Step-by-step explanation:

correct on quiz

4 0

3 years ago

Read 2 more answers

Solve 3x2−3x=0 by factoring.

o-na [289]

Answer:

3x(x-1) = 0

x= 1

Step-by-step explanation:

is 3x2 supposed to mean 3x^2 i dont know

5 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

4 years ago

Is this graph odd, even , or neither Drop answers pls

valina [46]

Answer:

even

Step-by-step explanation:

6 0

3 years ago

A right triangle has legs measuring 4.5 meters and 1.5 meters. The lenghts of the legs of a second triangle are proportional to

Eduardwww [97]

We know that
in the first triangle
the ratio of the legs are
4.5/1.5-----> 3

then
case A) 6 m and 2 m ------> ratio=6/3----> 3
so
the legs of a second triangle are proportional to the lengths of the legs of the first triangle

case B) 8 m and 5 m ------> ratio=8/5---->1.6
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case C) 7 m and 3.5 mm ------> ratio=7/3.5---->2
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case D) 10 m and 2.5 m ------> ratio=10/2.5---->4
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case E) 11.25 m and 3.75 m ------> ratio=11.25/3.75---->3
so
the legs of a second triangle are proportional to the lengths of the legs of the first triangle

the answer is
A) 6 m and 2 m
E) 11.25 m and 3.75 m

4 0

3 years ago