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2 years ago

15

The entire world's population could fit inside Los Angeles

2 answers:

jolli1 [7]2 years ago

8 0

I’m pretty sure that’s is not true.

7nadin3 [17]2 years ago

3 0

Answer:

Nah, I don't think so. But if that real, then woah

You might be interested in

Show answer in lowest terms. 3 1 /4 - 1 3/16=

SpyIntel [72]

3 - 1 = 2
1/4 - 3/16 = 4/16 - 3/16 = 1/16

answer is : 2 1/16 (in mixed number) or 33/16 (in improper fraction)

4 0

2 years ago

Which expression is equivalent to -2(5x – 0.75)?

katen-ka-za [31]

Answer:

0 - 10x + 1.5

Step-by-step explanation:

-2(5x - 0.75)

Expand the bracket

-10x + 1.5

0 - 10x + 1.5 is the equivalent expression

3 0

2 years ago

Read 2 more answers

A room has the area of 60 square meters and perimeter of 32 meters .What are the length and width of the room?

ad-work [718]

I am sorry but i don't no

4 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

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

7 0

2 years ago

Iona wrote out the description of each step for her multiplication of the binomial and trinomial (2x – 3)(5x2 – 2x + 7)

Olin [163]

Steps for multiplying the given binomial and trinomial $(2x-3)\times (5x^{2}-2x+7)$ are:

Step 1: $2x\times (5x^{2}-2x+7)-3 \times (5x^{2}-2x+7)$

Step 2: $10x^{3}-4x^{2}+14x-15x^{2}+6x-21$

Step 3: $10x^{3}-19x^{2}+20x-21$

7 0

2 years ago

Read 2 more answers