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Shalnov [3]
3 years ago
13

I need help with 8 and 9

Mathematics
1 answer:
Degger [83]3 years ago
8 0
3/4 is real & rational
-12/3 is real, rational, whole number & integer
0.345345345 is real and rational
-0.647349024 is irrational

Why?

3/4 is rational because it can be written as a fraction. Anything that's rational automatically is also considered a real number.

-12/3 is -4. -4 is an integer (a whole number) which automatically makes it real and rational.

0.345345345 is real because its a repeating decimal. If it wasn't repeating then it wouldn't be real and it wouldn't be rational.

-0.647349024 is just irrational cause its not a repeating (or terminating) decimal. :)
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5/6

Step-by-step explanation:

5/1 cakes shared by 6/1 people. To divide, we can freeze flip and multiply the fractions to get to 5/1 * 1/6 = 5/6 cakes per person. Let me know if this helps.

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A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable
Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. 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.

In this problem we have that:

n = 14, p = 0.3.

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

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

P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066

P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014

P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002

P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024

P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002

P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082

0.0082 = 0.82% probability that he will pass

6 0
3 years ago
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