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Nesterboy [21]
4 years ago
10

It is possible to construct a cube of twice the volume of a given cube using only a straightedge and compass.

Mathematics
2 answers:
Oxana [17]4 years ago
8 0

No, it is not.

That specific question is named the Delian Problem, where supposedly, Apollo, the Greek god, had asked for his altar to be doubled in size to stop a plague going around. The builders only had a compass and straightedge, but they couldn't figure out how to double the volume. The reason they couldn't was because there was no way to construct multiples of \sqrt[3]{2a}, which was what was needed to double the volume.

This problem is one called the impossible problems from antiquity, or problems the ancient Greeks could not solve with only a compass and a straightedge. Another example of one of these problems is how to trisect an angle.

Likurg_2 [28]4 years ago
8 0

Answer:NO

Step-by-step explanation:

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You’ve bought a half-dozen (six) eggs from the store but you forgot to check them first! The probability that no eggs are broken
GREYUIT [131]

Answer:

a)

P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4

P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960

P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633

P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359

P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044

P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003

P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001

b) 56.77% probability that an even number of eggs is broken.

c)

Expectation: 0.8496

Variance: 0.7293

Step-by-step explanation:

For each egg, there are only two possible outcomes. Either it is broken, or it is not. The probability of an egg being broken is independent from other eggs. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

There are 6 eggs

So n = 6

The probability that no eggs are broken is 0.4.

This means that P(X = 0) = 0.4. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.p^{0}.(1-p)^{6}

(1 - p)^{6} = 0.4

Taking the sixth root from both sides of the equality

(1 - p) = 0.8584

p = 0.1416

Each egg has a 0.1416 probability of being broken

(a) Write out the pmf of X.

Probability of each value, from 0 to 6

P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4

P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960

P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633

P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359

P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044

P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003

P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001

(b) Compute the probability that an even number of eggs is broken.

0, 2, 4 or 6

P = P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) = 0.4 + 0.1633 + 0.0044 + 0.00001 = 0.5677

56.77% probability that an even number of eggs is broken.

(c) Compute the expectation and variance of X.

Expectation:

E(X) = np = 6*0.1416 = 0.8496

Variance:

V(X) = np(1-p) = 6*0.1416*0.8584 = 0.7293

7 0
4 years ago
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