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Alexxandr [17]
1 year ago
6

Can anybody help me?

Mathematics
1 answer:
Kruka [31]1 year ago
4 0

f(-2) = 2(-2)² - 8(-2) + 6 = 30

f(-1) = 2(-1)² - 8(-1) + 6 = 16

f(0) = 2(0)² - 8(0) + 6 = 6

f(1) = 2(1)² - 8(1) + 6 = 0

f(2) = 2(2)² - 8(2) + 6 = -2

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Aleksandr [31]
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X ints = (-1,0) , (3,0)

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Step-by-step explanation:

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erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used 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.

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And p is the probability of X happening.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

p = \frac{4}{6} = 0.6667

10 normal six sided dice are thrown.

This means that n = 10

Find the probability of obtaining at least 8 failures.

This is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

So

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

P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951

P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867

P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174

Then

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992

0.2992 = 29.92% probability of obtaining at least 8 failures.

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