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2 years ago

R = {(-3, -2), (-3, 0), (-1, 2), (1, 2)}

Mathematics

1 answer:

vagabundo [1.1K]2 years ago

3 0

<h3>Answer: -3, -1, 1</h3>

Explanation:

The domain is the set of all allowed x values in a relation. We simply list each unique x coordinate of the (x,y) points given to form the domain.

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Beth is taking out a loan to purchase a new home. She is financing $50,000 for 25 years at an interest rate of 14.25%. What is h

kozerog [31]

The present value (PV) of a loan for n years at r% compounded t times a year where there is equal P periodic payments is given by:

$PV=P\left( \frac{1-\left(1+ \frac{r}{t} \left)^{-nt}}{ \frac{r}{t} } \right)$

Given that <span>Beth is taking out a loan of PV = $50,000 to purchase a new home for n = 25 years at an interest rate of r = 14.25%. Since she is making the payment monthly, t = 12.

Her monthly payment is given by:

$50,000=P\left( \frac{1-\left(1+ \frac{0.1425}{12} \right)^{-25\times12}}{ \frac{0.1425}{12} } \right) \\ \\ =P\left( \frac{1-(1+0.011875)^{-300}}{ 0.011875 } \right)=P\left( \frac{1-(1.011875)^{-300}}{ 0.011875 } \right) \\ \\ =P\left( \frac{1-0.028969}{ 0.011875 } \right)=P\left( \frac{0.971031}{ 0.011875 } \right)=81.770994P \\ \\ \therefore P= \frac{50,000}{81.770994} =\$611.46$

Therefore, her monthly payment is about $611.50
</span>

3 0

4 years ago

12. Find the measure of the largest angle in the triangle shown.

Art [367]

Step-by-step explanation:

all angles of a triangle sum to 180

180 = 15x + 24x +21x

180=60x

solve for x

then substitute that for 24(x) to get the angle

7 0

3 years ago

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taurus [48]

C is the answer to your question

5 0

3 years ago

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The school can food drive has collected an average of 171 cans of food per month from january to june. the balance at the end of

Nat2105 [25]

Answer:

1002

Step-by-step explanation:

171x6=1026

171×2=342

1026-342=684

684+318=1002

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