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3 years ago

Which graph has a coefficient correlation r closeest to -0.95

Mathematics

2 answers:

V125BC [204]3 years ago

6 0

Graph b has a correlation closest to -0.95

Maksim231197 [3]3 years ago

4 0

Answer:

Graph has a coefficient correlation r closest to -0.95 is the b).

Step-by-step explanation:

The correlation refers to the slope of a function and the sign indicates the direction to be linear the correlation informs us that it is inversely proportional

Taking some number to realice

Y axis maximum 5

X axis maximum 5

Notice 5 is where cut axis y so b=5

Equation:

$y=mx+b\\y=-0.95x + 5$

Replacing:

x1=1.25; x2=2.5; x3=3.8;

So:

(1.25, 3.8); (2.5, 2.625); (3.8, 1.39)

You might be interested in

Janet and her brother John went out to eat. Both paid the same amount. Janet got three burgers and left a $10 tip. John bought 4

lianna [129]

Answer:

Cost for 1 burger = 5 $

Amount spent by Janet = 25$

Amount spent by John = 25$

Explanation:

Let the cost for burger be p.

Janet got three burgers and left a $10 tip

Amount spent by Janet = 3 p + 10

John bought 4 burgers, but only left a $5 tip

Amount spent by John = 4 p + 5

Both paid the same amount.

So, we have 3 p + 10 = 4 p + 5

=> p = 5 $

So, cost for 1 burger = 5 $

Amount spent by Janet = 3 x 5 + 10 = 25$

Amount spent by John = 4 x 5 + 5 = 25$

6 0

2 years ago

A courtyard has three walls with widths

wariber [46]

Using the greatest common factor, it is found that the greatest dimensions each tile can have is of 3 feet.

---------------------------

The widths of the walls are of <u>27 feet, 18 feet and 30 feet.</u>
<u>The tiles must fit the width of each wall</u>, thus, the greatest dimension they can have is the greatest common factor of 27, 18 and 30.

To find their greatest common factor, these numbers must be factored into prime factors simultaneously, that is, only being divided by numbers of which all three are divisible, thus:

27 - 18 - 30|3

9 - 6 - 10

No numbers by which all of 9, 6 and 10 are divisible, thus, gcf(27,18,30) = 3 and the greatest dimensions each tile can have is of 3 feet.

A similar problem is given at brainly.com/question/6032811

8 0

3 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

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 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"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

I dont know the x of 30-2x=22

zloy xaker [14]

Answer:

Step-by-step explanation:

30-22=2x

8=2x

8/2=x

3 0

2 years ago

Read 2 more answers

HELP ASAP! Relations and Functions!!! Giving brainliest!!!

Leviafan [203]

Answer:

Step-by-step explanation:

Neither are a function because they do not pass the vertical line test.

4 0

3 years ago