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Andre45 [30]
2 years ago
15

Please help me with this I posted the pic

Mathematics
2 answers:
nikklg [1K]2 years ago
8 0
The answer is b, (-√3 , -1)
sergeinik [125]2 years ago
7 0
It’s f. my guy...........................

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A company charges $7 for a T-shirt and ships any order for $15. A school principal ordered a number of T-shirts for the school s
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7x + 15 = 1520
It’s $7 times the unknown number of shirts (x). Plus $15 to ship. All has to equal the total of $1,520.
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3 years ago
What is the median of this set of data?<br> 1, 2, 5, 6,9
Aleksandr-060686 [28]

Answer:

5

Step-by-step explanation:

median = (n+1)/2 th item

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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
3 years ago
I NEED HELP ASAP Catherine finds that in a randomly selected sample of 500 families in her home town, 45% own at least one compu
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A is the correct answer
6 0
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Read 2 more answers
Solve the inequality -6c&lt; -12
sweet [91]

Answer: c<2

Step-by-step explanation:

-6c<-12

c<-12/-6

c<2

7 0
3 years ago
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