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2 years ago

Raphael paid $656 for a camera during a 20% off sale. What was the camera's regular price?

1 answer:

8 0

$820.

If you divide 656 by 8 (since it's 80% of the total price), you get 82.
Multiplying 82 by 10 would equal the original price.

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A deli sells 2 hot dogs for $2.50. What is the constant of proportionality of dollars per hot dog?

attashe74 [19]

Answer:

$1.25 per hot dog

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Louis eat 6/8 of a pizza what fraction of the pizza is left over

arlik [135]

1 whole pizza is equal to 8/8. If Louis eats 6/8 then there should be 2/8 left, if your teacher wants it simplified then you will divide both the numerator and the denominator by 2 giving you a fraction of 1/4.

8 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

Factor Completely:<br> 4x²+8x

Sidana [21]

The answer is 4x(x+2).

8 0

3 years ago

Read 2 more answers

Which value of x is in the solution set of -3X+8≥14? ) A. -3 B. -1 C. 0 D. 3

romanna [79]

Answer:

Step-by-step explanation:

Solving for x:

-3x + 8 ≥ 14

-3x ≥ 6

x ≤ -2

Notice the sign gets reversed when you divide or multiply by a negative number.

The only option less than -2 is -3, A.

Here's a tip: on multiple choice questions, sometimes it's faster to try each answer instead of solving the equation. This can save time on tests like the SAT or ACT.

Let's try it:

-3(-3) + 8 = 9 + 8 = 17

-3(-1) + 8 = 3 + 8 = 11

-3(0) + 8 = 0 + 8 = 8

-3(3) + 8 = -9 + 8 = -1

So you can see the answer is A.

7 0

2 years ago