All categories

3 years ago

4, 4, 5, 7, 10, what's the next number in the pattern

Mathematics

2 answers:

Anna007 [38]3 years ago

8 0

Answer:

Step-by-step explanation:

sladkih [1.3K]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Which statement is true?

sveticcg [70]

Answer:

3rd

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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

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

7 0

3 years ago

Y=-x+7

Allisa [31]

Question 1:
y=-x+7
y=2x-2

(-x+7)=2x-2
-x+7+x=2x-2+x
7=2x-2+x
7+2=2x-2+x+2
9=3x
9/3=3x/3
3=x

y=-x+7
y=-(3)+7
y=4

Question 2:
y-1=2x
y+2x=5

y-1=2x
y-1+1=2x+1
y=2x

(2x+1)+2x=5
2x+1+2x=5
4x+1=5
4x+1-1=5-1
4x=4
4x/4=4/4
x=1

y-1=2(1)
y-1+1=2+1
y=3

Answer: x=1 y=3

6 0

3 years ago

Describe how to create a solid model of a simple coffee mug using sweep and other constructive solid geometry operations

anzhelika [568]

Answer:

Step-by-step explanation:

We are going the use the revolve constructive geometry operation

Step one:

Knowing fully well that the simple mug has a cylindrical shape, and all preliminary settings are in place, like the units may be in inches or mm.

We first need to draft out the section of the mug, this will look like an "L" shape with thickness set to whatever dimension you may have in mind.

The next in line is to create an axis using a construction line parallel to the L shape.

Then use the revolve tool to create the cylindrical shape by selecting the axis as the reference of rotation and the angle of rotation set to 360 degrees.

5 0

3 years ago

Add It!<br> 0.1496 + 0.7 =

Readme [11.4K]

Answer:

Step-by-step explanation:

0.8496

7 0

3 years ago

Read 2 more answers