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pickupchik [31]
2 years ago
15

Justin builds a triangular fenced area for his dog called Triangle ABC. If AB = 17 ft and BC = 8ft, and angle C is a right angle

, which of the following are true? Select all that apply.
ロThe missing side measures 12.5 ft.
ロ The missing side of the fence measures 15 ft.
ロ The perimeter of the fence is 40 ft.
ロ The perimeter is 37.5 ft.
ロ The area inside the fenced in section is 60 sq ft
ロ The area inside the fenced section is 68 sq ft.
Mathematics
1 answer:
Anika [276]2 years ago
8 0

The missing side of the fence measures 15 ft

The perimeter of the fence is 40 ft

The area inside the fenced in section is 60 ft²

The fence is triangular and a right triangle.

<h3>Right triangle:</h3>

A right triangle has one of its angle as 90 degrees. Therefore, the sides can be found using Pythagoras theorem.

c² = a² + b²

17² - 8² = b²

b = 15 ft

AC = 15 ft

perimeter of the triangle = 15 + 8 + 17 = 40 ft

Area of the triangle = 1 / 2 × 15 × 8 = 60 ft²

Therefore,

The missing side of the fence measures 15 ft

The perimeter of the fence is 40 ft

The area inside the fenced in section is 60 ft²

learn more on right angle triangle here; brainly.com/question/2192748

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On a certain​ exam, Tony corrected 20 papers and found the mean for his group to be 70. Alice corrected the remaining 10 papers
son4ous [18]

Answer:

The mean of the combined group is 76.67 ≈ 77

Step-by-step explanation:

  1. The sum of the grades in Tom's group is 20×70 = 1400
  2. The sum of grades in Alice's group is 10×90=900
  3. The sum of gradees in the combined group is =1400+900=2300
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where

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  • 30 is the total number of people in the group
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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 =
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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

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

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

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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}

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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<u>Equation</u>:

\Rightarrow \sf a - a1 = m(n - n1)

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\Rightarrow  \sf a = n - 7 + 10

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3 years ago
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