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Answer:
probability of an electric power outage shipboard is 0.075 or 7.5 %
Step-by-step explanation:
Given the data in the question;
Probability of engine malfunction = 25% = 0.25
Probability of generator malfunction = 10% = 0.1
Probability of generator being repaired = 50% = 0.5
probability of an electric power outage shipboard = ?
To get the probability of an electric power outage shipboard, we use the expression;
probability of an electric power outage = 2 × p( Probability of generator malfunction ) × p( Probability of generator being repaired ) × ( 1 - p( Probability of engine malfunction ) )
so we substitute in our given values;
probability of an electric power outage = 2 × 0.1 × 0.5 × ( 1 - 0.25 )
probability of an electric power outage = 2 × 0.1 × 0.5 × 0.75
probability of an electric power outage = 0.075 or 7.5 %
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.