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3 years ago

there are 12 actors auditioning for 5 different roles in a new play. how many different cast are possible

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

3 0

Answer:

95,040 different cast are possible

Step-by-step explanation:

In this case the order in which the actors are selected for the 5 different roles is important. Therefore we use the permutations formula.

$nPr =\frac{n!}{(n-r)!}$

Where n is the number of items you can choose and choose r from them.

In this case $n = 12$ and $r = 5$

$12P5 =\frac{12!}{(12-5)!}$

$12P5 =\frac{12!}{7!}$

$12P5 =95040$

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Six stand-up comics, A, B, C, D, E, and F, are to perform on a single evening at a comedy club. The order of performance is

kiruha [24]

Answer:

a) 1/6

b) 1/36

c) 1/720

d) 1/3

Step-by-step explanation:

a) Any of the six comics can perform in the fourth place, so there is one chance in six that Comic F is the one that performs fourth.

$P(F=4th)=1/6$

b) In this case, we have two conditions. Both are independent of each other, so the probability of both happening is the product of the probabilities of each happening individually:

$P(D=4th \&B=2nd)=P(D=4th)\cdot P(B=2nd)=(1/6)\cdot(1/6)=1/36$

c) This combination is one in all possible orders of perform. The amount of combinations of orders is n!=6!=720 possible combinations. So the probability of this specific order is:

$P(F, D, B, A, C, E)=1/720$

d) In this case, of the six possible comics performing in the last place, we calculate the probability of 2 of them being in that place. So the probability is:

$P(C=6th\,or\, E=6th)=2/6=1/3$

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3 years ago

7x(7) + 7(2) = 7(x+2)

Rudik [331]

Answer: One

Step-by-step explanation:

7x(7) + 7(2) = 7(x + 2)

49x + 14 = 7x + 14

49x = 7x

x = 0

7 0

2 years ago

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Gary is buying a $1,250 computer on the installment plan you makes a down payment on of $150 she has to make monthly payments of

Nataliya [291]

1250-150=1100
2 1/2 years = 30 months
30*48.25 = 1447.5 - 1100 = $<span>347.5 is you Answer</span>

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3 years ago

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Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,

Jet001 [13]

Answer:

$f(x,y,z)=ye^{xz}+C$

Step-by-step explanation:

We can write the given expression as :

$\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k$

As given, f = ∇f.

∇f = $\dfrac{\partial f}{\partial x}i$ + $\dfrac{\partial f}{\partial y}j$ + $\dfrac{\partial f}{\partial z}k$

We can write the partial derivative with respect to x, y and z.

$\dfrac{\partial f}{\partial x}=yze^{xz}$ ___(Equation 1)

$\dfrac{\partial f}{\partial y}=e^{xz}$ ______(Equation 2)

$\dfrac{\partial f}{\partial z}=xye^{xz}$ ______(Equation 3)

Take equation 2 and integrate with respect to y,

$\dfrac{\partial f}{\partial y}=e^{xz}$

$f(x,y,z)=ye^{xz}+a(x,z)$ ----------Equation 4

Derivate both sides w.r.t x , we get :

$\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}$

$\dfrac{\partial a}{\partial x}=0$

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

$f(x,y,z)=ye^{xz}+b(z)$

take derivative wrt z

$\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}$

we can take here:

$\frac{db}{dz} = 0$

integrate:

$\int\ {\frac{db}{dz} } \, =\int0$

b(z) = C

The function can be written as :

from equation 4 :

$f(x,y,z)=ye^{xz}+C$

Where C is a constant.

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2 years ago

7 – (5t - 13) = -25<br><br> i need help with this question

Mazyrski [523]

Answer:

t=9

Step-by-step explanation:

Solve -

Distribute :7-(5t - 13) = - 25

7 - 5t + 13 = - 25

Add the numbers :7 – 57 + 13 = -25

20-51 = -25

Rearrange terms :20-51 = -25

-51+ 20 = -25

Subtract 20 from both sides :-51 + 20 = -25

-51+20-20-25-20

Simplify :Subtract the numbers

-51+ 20-20-25-20

-5:=-25-20

Subtract the numbers

-51 = -25-20

-51 = -45

Divide both sides by the same factor : -51 = -45 -5t -5 -45- -5 =

Simplify : Cancel terms that are in both the numerator and denominator :: -5t - = -45 - -5

t = - 45 - -5

Divide the numbers :

t = - 45 - -5 ” t=9 ” here’s the answer

4 0

2 years ago