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2 years ago

I need to know how to describe the parts of the expression (3x-2)

Mathematics

2 answers:

zalisa [80]2 years ago

5 0

Answer:

If the arrow is pointing to the -2, that’s the constant.

If the arrow is pointing at x, that’s the variable.

If the arrow is pointing at 3, that’s the coefficient.

stepladder [879]2 years ago

5 0

Answer:

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I need help on how to solve this problem?

Damm [24]

X = -2
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2 years ago

Read 2 more answers

I don’t know how to solve this

V125BC [204]

Assuming that both triangles are an exact copy of one another, it is safe to assume that 3y-7 is equal to 41. Set up an equation

3y-7=41

Add 7 to both sides

3y=48

Divide both sides by 3

y=16

Now to find PN.

Based on what we know, we can assume that MP = PN. Let's make some equations!

MP = 17x-8 PN = 11x+4

17x-8 = 11x+4

Subtract 11x from both sides

6x-8 = 4

Add 8 to both sides

6x = 12

Divide by 2

x=2

Substitute 2 in for x in the equation for PN

11(2)+4

Multiply 11 by 2

22+4 = 26

PN = 26

7 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

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2 years ago

Please help me with this math problem

vfiekz [6]

The function is decreasing on the interval (- infinity, + infinity).

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2 years ago

14 POINTS ASAP WHAT IS THE MEASURE OF ANGLE A

sashaice [31]

Wow...we have lots of numbers to go through. we know that a 6 sided figure is a hexagon and the interior angles add up to be 720°. so.....

∠A + ∠B + ∠ C + ∠D + ∠E + ∠F = 720°
(x - 60) + (x - 40) + 130 + 120 + 110 + (x - 20) = 720
3x + 240 = 720 (combined all like terms)
3x = 480 (subtracted 240 from both sides)
x = 160 (divided both sides by 3)

put the value of x into ∠A (x - 60) = 160 - 60 = 100

∠A = 100°

7 0

2 years ago