Ask question

Ask question

All categories

3 years ago

8

Your favorite radio station is having a contest! The host poses a question to the listeners. If the caller answers correctly, he

or she wins the prize money. If the caller answers incorrectly, $30 is added to the prize money and the next caller is eligible to win! The contest starts with a $100 prize for the first caller.
Part A: The first 4 callers have all missed the question. How much money is the 5th caller eligible to win? Show your work or explain your thinking.

Part B: You are the 12th caller to call in, and you win the prize! How much money will you win?

2 answers:

Arisa [49]3 years ago

5 0

Answer:

A:150

B: 360 if all before are wrong

Maslowich3 years ago

3 0

Answer:

A 150

Step-by-step explanation:

B- 360 if all before are wrong

You might be interested in

A function is a type of ___________.

miv72 [106K]

Answer:

A function is a type of <em><u>vertical line </u></em><em><u>test.</u></em>

5 0

2 years ago

Read 2 more answers

A hockey player takes 150 shots each week. Sixty percent of these shots go in. How many shots do NOT go in?

Viefleur [7K]

60 percent of 150 is 90. So that means 60 shots do not go in.

4 0

3 years ago

Lauren is selling tickets for her class play. She has already sold twice as many tickets this year as she did last year. If she

Aleksandr [31]

Lauren needs to sell 6 more tickets to reach her goal. 30- (12x2) 12x2=24 30-24=6

5 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

An apple orchard sold 155 apples in September in November it sold 17 times as many apples as it did in September how many apples

borishaifa [10]

9514 1404 393

Answer:

2635

Step-by-step explanation:

17 times the September number is ...

17×155 = 2635 . . . . apples sold in November

8 0

3 years ago