In a certain four-engine vintage aircraft, now quite unreliable, each engine has a 15% chance of failure on any flight, as long as it is carrying its one-fourth share of the load. But if one engine fails, then the chance of failure increases to 35% for each of the other three engines. And if a second engine fails, each of the remaining two has a 45% chance of failure. Assuming that no two engines ever fail simultaneously, and that the aircraft can continue flying with as few as two operating engines, find the probability of exactly one engine failure.

0.0975 or 9.75%

Step-by-step explanation:

Assuming that no engines fail simultaneously, the probability that exactly one engine fails is the probability that the first engine fails (15%) multiplied by the probability that the second engine does not fail (100% - 35%):

The probability of exactly one engine failure is 0.0975 or 9.75%

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

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step-by-step explanation: