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3 years ago

Please help me answer this question quickly as possible

Mathematics

1 answer:

GuDViN [60]3 years ago

8 0

Answer:

the blanks for 54 are: 1,2,3,6

the blanks for 42 are: 1,2,3

HCF: 3

Step-by-step explanation:

the factors of each number are:

1 54 1 42

2 27 2 21

3 18 3 14

6 9

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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

If the equation of the line is y=1/2x-3, find the x-coordinate of the Lundy whose y coordinate is 5.

Anna71 [15]

Y=1/2x -3 find the x coordinate of the point whose y

coordinate is 5

y = 5. Write the equation as:

1/2F2x - 3 = 5

multiply both sides by 2, and you have

x - 6 = 10

x = 10 + 6

x - 16

Check solution in original equation, replace x with 16

y = 1/2F2(16) - 3

y = 8 - 3

y = 5

5 0

3 years ago

Explain why the graph below does not represent a direct variation

Nitella [24]

Answer:

C. The line does not go through the origin.

Step-by-step explanation:

We have been given a straight line graph that doesn't pass through origin (0,0) and has y-intercept at (0,3).

Now we need to decide why that graph doesn't represent a direct variation.

We know that graph represents variation when it passes through origin. But since given graph doesn't passes through origin so it doesn't represent the direct variation.

Correct choice is :

C. The line does not go through the origin.

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Ted pays $35 per hour for scuba class. Identify the constant of proportionality that relates his scuba class charges (y) to the

Naddik [55]

The answer would be D

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Is the following sequence arithmetic? -14, 86, 186, 286, ... A.No B.Yes