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11111nata11111 [884]
3 years ago
15

Hi I need help with this problem. But it has to be shown with work. Does anyone know?

Mathematics
2 answers:
SpyIntel [72]3 years ago
8 0

Answer:

0.4 or 2/5???

Step-by-step explanation:

I believe this is a proportion/ratio problem.

Think of it like this:

x/6 = 10/150

To solve for x, you must cross multiply:

150x = 6*10 (150x... DON'T write 150*x, that is nonsense)

150x = 60

     x = 60/150  (you divide because you are actually dividing 150 on both sides and doing the opposite operation of multiplication)

     x = 6/15

     x = 2/5 or 0.4  (I simplified it)

Hopefully this helped? I can't seem to find a whole number answer... so the only options I could do are 2/5 or 0.4

Damm [24]3 years ago
5 0

Answer:

its is 90

Step-by-step explanation:

6x10=60    150-60=90

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A computer is normally $550 but is discounted to $385. What percent of the original price does Mark pay?
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Answer:

Step-by-step explanation:

70% all you have to do is divide 355/550like that and times the answer by 100

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Read 2 more answers
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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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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If we want to use addition, add the opposite

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