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USPshnik [31]
3 years ago
12

A roulette wheel contains the integers 1 through 36, 0, and 00. suppose that you spin the wheel 6 times and that each time you b

et on a single number. what is the probability (rounded to the nearest hundredth) that you win on at least one bet?
Mathematics
1 answer:
kompoz [17]3 years ago
5 0
The probabability of winning on at least 1 bet is equal to 1 less the probaility of not winning on either of the 6 bets.

The probability of not wining on any bet is independent of winning or not winning on any of the bets, so the combined probability is calculated as the product of each individual probability.

Each indivitual probability of not winning the is:

(number of not winning outcomes) / (number of possible outcomes) = 37 / 38.

Then, the combined probability of not winning the six times is: (1/38)*(37/38)*(37/38)*(37/38)*(37/38)*(37/38) =(37/38)^6

Therefore, the probability of winning at least one bet is:

= 1 - (37/38)^6 ≈ 1 - 0.973684 ≈ 0.03.

Answer: 0.03.
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I need help on this please
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Answer:

0.27

Step-by-step explanation:

You take 1.75, and subtract 1.48 from it.     1.75

                                                                    <u>- 1.48</u>

                                                                      0.27

4 0
3 years ago
30% of 60 is <br><br> 30%of ____ is 60
ololo11 [35]
You know that 3 times 2 is 6 so 30'times 20 equals 60
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Add (-x^2-2x+6)+(-9x^2-7x+6)
77julia77 [94]

(-x²-2x+6)+(-9x²-7x+6) = -x²-2x+6-9x²-7x+6 = -10x² - 9x + 12

5 0
3 years ago
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

------------------

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

=============================================

Problem 6

<h3>Answer:  -3i</h3>

------------------

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

------------------

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

------------------

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
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katovenus [111]

Answer:

The answer is (A)

Step-by-step explanation:

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