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Naily [24]
2 years ago
8

Will give BRAINLIST

Mathematics
1 answer:
Kipish [7]2 years ago
4 0

Answer:

Thirty two! 32 is the answer.

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Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
How do you do to the 2nd power
Ray Of Light [21]
Something to the second power is simply that number multiplied by itself once. The third power is something multiplied by itself 2 times, and so on
4 0
3 years ago
Julie earned 66 on the test. Describe how the scores of Julie’s classmates compared to Julie’s score.
Ne4ueva [31]
We would need the data of their classmates scores to complete this answer
4 0
3 years ago
1) How is circumference different than area?
Fiesta28 [93]

Answer:

0.8? maybe

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
A spiral staircase turns as it rises 10 feet. The radius of the staircase is 3 feet. What is the number of feet in the length of
jenyasd209 [6]

The number of feet in the length of the handrail is 17.3 feet.

<h3>What is a Pythagorean Theorem?</h3>

Pythagorean theorem, is a geometric theorem that the total of a squares on the sides of a right triangle equals the square upon that hypotenuse (a side opposite its right angle)—or, using familiar algebraic notation, the sum of squared on the hypotenuse equals the square also on hypotenuse.

H² = B² + P²

A spiral would be a shape that wraps around and around with each curve being higher or lower than the one before it.

Now, according to the question;

The simplest way to understand this is to assume that we may "unfold" this spiral to form a rectangle.

Its bottom part of rectangle is simply an arc length of a 3 ft radius circle spun around 270°= (3/4) of a full revolution.

Thus,

= (3/4)× (2\pi) ×(3)  

=  (9/2)×\pi ft

=   4.5×\pi  ft

The height of the staircase will be the side of rectangle = 10ft.

And the rectangle's diagonal will correspond to the length of a railing.

To solve this, we can apply the Pythagorean Theorem.

Handrail length = √ [ (4.5 ×\pi)² + 10² ]  

Handrail length ≈  17.3 ft

Therefore, the the number of feet in the length of the handrail is 17.3 feet.

To know more about Pythagorean Theorem, here

brainly.com/question/343682

#SPJ4

The complete question is-

A spiral staircase turns 270° as it rises 10 feet. The radius of the staircase is 3 feet. What is the number of feet in the length of the handrail?

7 0
2 years ago
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