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IceJOKER [234]
4 years ago
9

Which system of linear equations had a solution of (-2, 1) ?

Mathematics
1 answer:
Oksana_A [137]4 years ago
7 0
You are correct, the answer is b
You might be interested in
What is the missing term of ..9, __ , 19
VARVARA [1.3K]

Answer:

The answer could be anything, but if it's an arithmetical sequence, then it's 14.

Step-by-step explanation:

I'm assuming that the added numbers always stay the same, meaning it's linear. If you add 5 to 9, you get 14. If you add 5 again, you get 19. Another way of doing this is (19-9)/2 = 5, then 9+5=14.

6 0
3 years ago
X^2 is to 441 as x^3 is to
labwork [276]

X=perfect square of 441

= 21

X^3 = 21^3

=9261

8 0
3 years ago
What is an expression equivalent to x - 3y + 4
Murljashka [212]

Answer:

Cannot be factored

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
Factor the expression below so it is in simplest form.<br> 8(3x + 5) + 14
shtirl [24]

Answer:

24x+54

Step-by-step explanation:

8(3x+5)= 24x+40

24x+40+14=24x+54

24x+54

8 0
3 years ago
Read 2 more answers
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