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Yanka [14]
2 years ago
7

How many ways are there to put 8 beads of different colors on the vertices of a cube, if rotations of the cube (but not reflecti

ons) are considered the same?
Mathematics
2 answers:
Delvig [45]2 years ago
6 0
What is being requested, if I'm not mistaken, is the number of permutations for placing each of the 8 beads on the vertices of the cubes;
In this case, we have 8 different beads and 8 possible locations for each of them;
So the number of permutations is:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
schepotkina [342]2 years ago
3 0

Actually, this is the correct answer.


Consider one vertex of the cube. When the cube is rotated, there are 8 vertices which that vertex could end up at. At each of those vertices, there are 3 ways to rotate the cube onto itself with that vertex fixed. So, there are a total of 8*3=24 ways to rotate a cube. There are 8! ways to arrange the beads, not considering rotations. Since the arrangements come in groups of 24 equivalent arrangements, the actual number of ways to arrange the beads is 8!/24=1680.

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The multiplication law is potentially helpful when we are interested in computing the probability of a. mutually exclusive event
Korvikt [17]

Answer:

b )  the intersection of two events

Step-by-step explanation:

Gary and Steve are both hosting . There are 50 buttons total, 15 buttons are blue and 27 buttons are red. Gary puts all of the buttons into a bag. Steve and Gary both want to wear red buttons.What is the probability ? To solve this problem, you need to understand the Multiplication Rule of Probability.This probability  means to find the probability of the intersection of two events, multiply the two probabilities.

Probability of two events occurring that is called intersection of two events. There are two different set of events , called independent and dependent events.

Independent events  event is not affected by a previous event.

A dependent event is when one event influences the outcome of another event . To find the intersection of two events, whether they are independent or dependent, multiply the two probabilities together.

3 0
2 years ago
Before the distribution of certain statistical software, every fourth compact disk (CD) is tested for accuracy. The testing proc
lilavasa [31]

Answer:

a) For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

P(R) =\frac{1}{4} =0.25

The failure probability for each program are given by:

P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932

So then the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

And if we use the fact that we have 4 possible test the true probability of interest would be:

P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017

b) p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))

And replacing we got:

p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484

c) From part a we now that the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

So then if we have 100 CDs the expected number of rejected Cd's are:

100*0.068= 6.8 \approx 7

Step-by-step explanation:

Part a

For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

P(R) =\frac{1}{4} =0.25

The failure probability for each program are given by:

P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932

So then the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

And if we use the fact that we have 4 possible test the true probability of interest would be:

P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017

Part b

For this case we want the probability that it failed program 2 or 3

So then we can find this probability like this:

p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))

And replacing we got:

p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484

Part c

From part a we now that the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

So then if we have 100 CDs the expected number of rejected Cd's are:

100*0.068= 6.8 \approx 7

3 0
2 years ago
Brainlest answer!!!!!!
Licemer1 [7]

Answer:

It can't be simplified

Step-by-step explanation:


3 0
2 years ago
Which of the following are correct statements concerning the roots of unity? Check all that apply.
mr Goodwill [35]

Answer:

C, B, and A

-Apex

6 0
2 years ago
Read 2 more answers
Mustafa’s soccer team is planning a school dance as a fundraiser. The DJ charges $200 and decorations cost $100. The team decide
irina1246 [14]

First, add the profit wanted with the costs.


1500 + 200 + 100 = 1800


The total amount wanted is 1800. Each student pays $5, and so you divide by 5


1800/5 = 360


360 students must show up.


The equation used is:


Let student = s


Total cost = 200 + 100

Charge = 5n

Profit = 1500


1500 = 5n - 300 (is equation)


Hope this helps

7 0
2 years ago
Read 2 more answers
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