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Anna007 [38]
2 years ago
9

. Quantas senhas com 4 algarismos diferentes podemos escrever com os algarismos 1, 2, 3, 4, 5, 6?

Mathematics
1 answer:
ruslelena [56]2 years ago
4 0

Answer:

We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

Step-by-step explanation:

We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.

Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.

Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.

  • Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).
  • Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).
  • Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).
  • Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).

So, the number of passwords with 4 different digits we can write = 6 \times 5 \times 4 \times 3  = 360 possibilities.

Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

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Please see the graph below

Step-by-step explanation:

The table is just selecting numbers for x and then solving for what y would be.  Once we have the ordered pairs (x,y)  We can graph the points and see the line.

3 0
1 year ago
2w(w2 + 3w - 5) + 3w3 + 2(w2 + 1)
marysya [2.9K]

Answer:

=5w3+8w2−10w+2

Step-by-step explanation:

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2

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Combine like terms

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7 0
2 years ago
You take a quiz with 6 multiple choice questions. After you studied your estimated that you would have about an 80 % chance of g
Arada [10]

Answer:

1.15%

Step-by-step explanation:

To get the probability of m  independent events you multiply the individual probability of each event. In this case we have m independent events, each one with the same probability, therefore:

p^{m}

0.8^{20} = 1.15\%

This is a particlar scenario of binomial distribution problem. So the binomial distribution questions are about the number of success of m independent events, where every individual event has the same p probability. In the question we have 20 events and each event has a probability of 80%. The binomial distribution formula is:

\binom{n}{k} * p^{k} * (1-p)^{n-k}

n is the number of events

k is the number of success

p is the probability of each individual event

\binom{n}{k} is the binomial coefficient

the binomial coefficient allows to find the subsets of k elements in a set of n elements.  In this case there is only one subset possible since the only way to get 20 of 20 correct questions is to getting right all questions (for getting 19 of 20 questions there are many ways, for example getting the first question wrong and all the other questions right, or getting second questions wrong and all the other questions right, etc).

\binom{n}{k} = \frac{n!}{k!(n-k)!}

therefore, for this questions we have:

\frac{20!}{20!(20-20)!} * 0.8^{20} * (1-0.8)^{0} = 1.15\%

4 0
2 years ago
What's (-1)³ simplfied?
klasskru [66]

Yo sup??

(-1)³=(-1)*(-1)*(-1)

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Hope this helps.

3 0
3 years ago
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Is it true or false lollll
Reika [66]

Answer:

I think true im sorry if its not

4 0
2 years ago
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