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tamaranim1 [39]

2 years ago

13

At a football stadium, 10% of the fans in attendance were teenagers. If there were 30 teenagers at the football stadium, what wa

s the total number of people at the stadium?
Insert the values given in the problem then scale up or down to find the missing value.

1 answer:

JulijaS [17]2 years ago

8 0

Answer:

i believe the answer to this is a total of 300 people because you multiply the ten percent into the 30 people which factors out some numerical values, so then your answer as i just said is 300 people in total attended the game.

Step-by-step explanation:

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Answer:

Minimum unit cost = 5,858

Step-by-step explanation:

Given the function : C(x)=x^2−520x+73458

To find the minimum unit cost :

Take the derivative of C(x) with respect to x

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Set = 0

2x - 520

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Hence, minimum unit cost will be :

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Put x = 260

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2 years ago

Two thirds of 12 minus d

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Step-by-step explanation:

2 /3 of 12 - d

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3 years ago

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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

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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

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

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

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

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

Eddie built the ramp shown to train his puppy to do tricks. What is the surface area of the ramp. The dimensions are 20,20,20,23

mamaluj [8]

Answer:

3775 in²

Step-by-step explanation:

The surface area of the ramp :

Area of rectangle = 20 * 23 = 460 in²

Area of rectangle = 20 * 23 = 460 in²

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Bottom rectangle = 60 * 23 = 1380 in²

For the trapezium: (front and rear)

1/2 (base 1 + base 2) * height

Base total = 20+ 20+ 20= 60

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Total surface Area :

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6 0

3 years ago