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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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Help!!!!<br>Given log7⁡3≈0.5646 and log7⁡16≈1.4248, evaluate the expressions.​
Alinara [238K]

Answer:

a)  0.356

b)  1.1397

Step-by-step explanation:

a) log₇2

  • log(2) / log(7)
  • 0.356

b) log₇ (¹⁴⁷/₁₆)

  • <u>log (¹⁴⁷/₁₆)</u>

          log (7)

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6 0
2 years ago
4. a) A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If t
Klio2033 [76]

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

a_{n}=a_{1}r^{n-1}

a_{1} = the first term of the series

r = the geometric ratio

a_{1} would represent the height at which the ball is first dropped. Therefore:

a_{1} = k

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

r=\frac{3}{4}

Our general term would be:

a_{n}=a_{1}r^{n-1}

a_{n}=k(\frac{3}{4}) ^{n-1}

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

k represents the initial height:

k = 235\ ft

n represents the number of times the ball bounces:

n = 6

Plugging this back into our general term of the geometric series:

a_{n}=k(\frac{3}{4}) ^{n-1}

a_{n}=235(\frac{3}{4}) ^{6-1}

a_{n}=235(\frac{3}{4}) ^{5}

a_{n}=55.8\ ft

a_{n} represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. ​

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

We already know these variables:

a_{1}= k = 235\ ft

r=\frac{3}{4}

n = 12

Therefore:

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }

S_{n}=(4)(235)(1-\frac{3}{4} ^{12})

S_{n}=910.22\ ft

8 0
2 years ago
Wilbur runs 3.2 km every morning how many feet are in 3.2 km given that 1 mile = 1.609 km and 1 mile = 5,280
Nonamiya [84]
1.609/5280 = 3.2/x....1.609 km to 5380 ft = 3.2 km to x ft
cross multiply because this is a proportion
(1.609)(x) = (5280)(3.2)
1.609x = 16896
x = 16896/1.609
x = 10,500 ft <===
5 0
3 years ago
Read 2 more answers
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