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Xelga [282]
2 years ago
11

Solve for m. -3/7m = 12 m= -28 m = -26 m = 26 m = 28 Please help

Mathematics
1 answer:
stellarik [79]2 years ago
4 0
The answer would be -28
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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
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0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from 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.

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This means that n = 15, p = \frac{1}{5} = 0.2

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P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

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

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

0.0111% probability that he answers at least 10 questions correctly

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