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Murrr4er [49]
3 years ago
11

Which system represents the graph?

Mathematics
2 answers:
adelina 88 [10]3 years ago
4 0

Answer:

im not sure but i think c or the third one down

Step-by-step explanation:

Morgarella [4.7K]3 years ago
3 0
It’s the last one!! hope it helps
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SOMEONE PLS HELP :(!!!!!!
diamong [38]
It would be C as the answers
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2 years ago
Winslow buys 1.2 pounds of bananas. The bananas cost $1.29 per pound. To the nearest cent, how much does Winslow pay for the ban
NNADVOKAT [17]
 1.2*1.29= $1.55
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4 0
3 years ago
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

8 0
3 years ago
Does the table below represent a proportional relationship?
Arturiano [62]
C. Yes, because all the rations of y to x are equal to 3 (: hope this helps
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3 years ago
Use the data below to determine how many days of class a student missed if he or she has a quarter grade of 40%.
Murljashka [212]

Answer:7 days

Step-by-step explanation:

8 0
3 years ago
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