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Sedbober [7]
3 years ago
13

Answer the question about the angles below.

Mathematics
1 answer:
kaheart [24]3 years ago
3 0
To find angle Y, take 180 degrees (linear pair) minus 35 degrees (angle X).
180-35=145 degrees
Angle Y=145 degrees
Angle W is also 145 degrees since it’s a vertical angle to Angle Y.
To find angle N, take 90 degrees (complementary angle) minus 55 degrees (angle M).
90-55=35 degrees
Angle N=35 degrees
Hope this helped!
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an equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each
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A 4 1/2 hamburger patty has 25 1/2 grams of protein,and 6 ounces of fish had 32 grams of protein
Savatey [412]

Answer: 4.5 oz/25.5 g = 1 oz/X g  

Step-by-step explanation: Crossed multiplies to solve for X.

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Given that triangle ABC is a right triangle, select a set of possible side length measures.
ad-work [718]

You didn't supply any rules or constraints so.... 1 in., 1 in., and 1 in.

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a^2 + b^2 = c^2

1^2 + 1^2 = 1^2 –this is true.

5 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
If we get the 100 from number 1 each year for ten years and invest each payment in an account that earns 8% how much will be the
blsea [12.9K]

Answer:

$1,448.66

Step-by-step explanation:

The future value of an annuity with yearly deposits 'P' at an interest rate of 'r' invested for 'n' years is determined by:

FV = P[\frac{(1+r)^n-1}{r}]

For P = $100, r = 0.08 and n = 10 years:

FV = 100[\frac{(1+0.08)^{10}-1}{0.08}]\\FV=\$1,448.66

The amount at the end of the ten years is $1,448.66

4 0
3 years ago
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