For this problem, assume 9 males audition, one of them being Winston, 5 females audition, one of them being Julia, and 6 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.

How many different ways can these roles be filled from these auditioners?

Available: 9M and 5F and 6C

Cast: 3M and 1F and 2C

As it is not ordered: C₉,₃ * C₅,₁ * C₆,₂

C₉,₃ = 9!/3!.6! = 84

C₅,₁ = 5!/1!.4! = 5

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₅,₁ * C₆,₂ = 84.5.15 = 6300

How many different ways can these roles be filled if exactly one of Winston and Julia gets a part?

2 options Winston gets or Julia gets it:

1) Winston gets it but Julia no:

8 male for 2 spots

4 females for 1 spot

6 children for 2 spots

C₈,₂ * C₄,₁ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₄,₁ = 4!/1!.3! = 4

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₄,₁ * C₆,₂ = 28.4.15 = 1680

2) Julia gets it but Winston does not

8 male for 3 spots

1 female for 1 spot

6 children for 2 spots

C₈,₃ * C₆,₂

C₈,₃ = 8!/3!.5! = 56

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₆,₂ = 56.15 = 840

1) or 2) = 1) + 2) = 1680 + 840 = 2520

What is the probability (if the roles are filled at random) of both Winston and Julia getting a part?

8 male for 2 spots

1 female for 1 spot

6 children for 2 spots

C₈,₂ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₆,₂ = 28.15 = 420

p = 420/6300 = 0.067

6 0

3 years ago

A store owner is interested in opening a second shop. She wants to estimate the true average daily revenue of her current shop t

Alex

Every confidence interval has associated z value. As confidence interval increases so do the z value associated with it.
The confidence interval can be calculated using following formula:
$\overline{x} \pm \frac{zs}{\sqrt n}$
Where $\overline{x}$ is the mean value, z is the associated z value, s is the standard deviation and n is the number of samples.
We know that standard deviation is simply a square root of variance:
$s=\sqrt{315900.20}=\$562.05$
The confidence interval of 95% has associated z value of <span>1.960.
</span>Now we can calculate the confidence interval for our income:
$3472.00\pm \frac{1.960\cdot 562.05}{\sqrt{60}}\\ \$3472.00\pm142.22$

6 0

3 years ago

Plz help me I've been dealing with this for 4 days.

Sloan [31]

Step-by-step explanation:

(qvp)5=4-5=54643434346

3 0

2 years ago