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3 years ago

What is the slope of 2y-2x=2

Mathematics

2 answers:

Anastaziya [24]3 years ago

6 0

Answer:

m = 1

Step-by-step explanation:

To easily find the slope of this equation, let's change it to slope-intercept form, y = mx + b.

2y - 2x = 2

Adding 2x to both sides, the equation becomes 2y = 2x + 2.

Dividing both sides by 2, the equation becomes y = x + 1.

The coefficient of x is the slope, and since there is no visible coefficient, that means the slope is 1 (AKA, x is being multiplied by 1).

vodomira [7]3 years ago

6 0

Answer:

Step-by-step explanation:

2y - 2x = 2

2y = 2x + 2

y = 2x/2 + 2/2

y = x + 1

Slope =1

{Compare with y =mx +b}

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Two cones have the same radii but one has a height four times larger than the other. How many small cones would be needed in ord

hram777 [196]

Since the radius is the same but the height changes, the volume would change by the multiple of the height.
In this case the volume of the larger cone would be 4 times as large as the smaller cone so you would need 4 small cones

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3 years ago

107=(9X-115) +(4x+27)

Cloud [144]

Answer: x=15

Step-by-step explanation:

Equation: 107=(9x-115)+(4x+27)

Step 1: 9x+4x-115+27=107

Step 2: 13x-115+27=107

Step 3: 13x-88=107

Step 4: 13x-88+88=107+88

Step 5: 13x=195

Step 6: 13x/13=195/13

x=15

6 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

Are the two triangles below similar??

kodGreya [7K]

Answer:

Yes

Step-by-step explanation:

If you find the value of all the angles, you'll find that both triangles' angles are 50, 35, and 90 degrees.

(all the angles in a triangle add up to 180)

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2 years ago

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balu736 [363]

Answer:

Step-by-step explanation:

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2 years ago

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