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What is the solution to P over 0.3 + 2 = 6? 2.4 1.2 13.3 26.7

Mathematics
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You almost have the right answer. This is an example of a binomial distribution problem. If an experiment can only have two outcomes (usually written "success" and "failure," but in this case E and "not E"), the probability of success (in this case E) is p, the probability of failure (in this case "not E") is q = 1 - p, and the experiment is repeated N times with the outcomes of each experiment being independent of each other, then the probability of exactly k success (and thus N - k failures) is: P(k successes) = C%28N%2C+k%29+%28p%5Ek%29+%28q%5E%28N-k%29%29+=+%28%28N%21%29%2F%28%28k%21%29%28N+-+k%29%21%29%29+%28p%5Ek%29+%28q%5E%28N+-+k%29%29
In this case, N = 6, k = 2, N - k = 4, p = 0.3, and q = 1 - 0.3 = 0.7. So we substitute into the formula: P(k successes) = +%28%286%21%29%2F%28%282%21%29%284%21%29%29%29+%280.3%5E2%29+%280.7%5E4%29+=+15+%2A+%280.3%5E2%29+%2A+%280.7%5E4%29+=+0.324135+ So your only mistake is leaving off the exponent on 0.7.
first do 6-2 then 4 times 0.3 thats how u find the value of p

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First you subtract 2 on both sides- P/0.3+2=4 -2 -2 ---------- That gives us P/0.3=2 Now we multiply 0.3 to both sides so we can cancel out the 0.3 under the P, and leave the p all by itself. |dw:1345216749247:dw| This finally gives us P=1.2 :D -Hope that helps :)
Thanks ^_^

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