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3 years ago

Leon drove 100 miles in 2 hours. Find the constant of variation.

Mathematics

2 answers:

Nikitich [7]3 years ago

7 0

The constant of variation is 50.
50 miles per hour.

Helga [31]3 years ago

3 0

To find the constant variation, solve for only one hour (in this case).

Divide 100 with 2

100/2 = 50/1

Each hour is 50 miles driven, so your constant variation is 50 mph

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Order from least to greatest 5/12,2/3,1/2,5/6,3/4

Effectus [21]

Answer:

Step-by-step explanation:

5/12

1/2

2/3

3/4

4 0

3 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

3 years ago

Ramona wants to buy cheese pizza for her family and friends gathering. She stop by nine wonder pizza and figures out that they h

Kryger [21]

Answer:

Large cheese

Step-by-step explanation:

The large cheese because it's two more dollars than the medium and she has a lot of people gathering

6 0

3 years ago

Read 2 more answers

I need help someone please

Andrew [12]

G(x)=2...............

5 0

3 years ago

1/2r -3= 3 (4-3/2r) solve for r

s2008m [1.1K]

1/2r -3= 3 (4-3/2r) is to be solved for r.

I'll begin by making the assumptions that by 1/2r you actually meant (1/2)r and that by 3/2r you actually meant (3/2)r. When in doubt, please use parentheses to make your meaning clear.

Thus, 1/2r -3= 3 (4-3/2r) becomes (1/2)r -3= 3 (4-(3/2)r ) .

Simplify this by multiplying all 3 terms by 2. Doing this will eliminate the fractions:

r -6 = 3 (4*2-(3)r ) or r - 6 = 24 - 9r

Now expand the right side, using the distributive property of

r - 3 = 24 - 9r

Regrouping so as to combine like terms:

10r = 30

Solving for r: r = 30/10 = 3

The value of r that satisfies this equation is 3

8 0

3 years ago