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Novosadov [1.4K]
3 years ago
7

PLEASE HELP FAST ILL MARK YOU BRAINLYEST

Mathematics
1 answer:
Alborosie3 years ago
4 0

Answer:

look at explanation (I go down the first column of scenarios and then move on to the right column)

Step-by-step explanation:

1) The first one will be same as original because the decrease and increase are the same, and happen right after each other so you get the same amount

2) Second one you get greater than original. You're doubling the initial value, and then decreasing by a smaller percentage, so you still get greater than your initial value

3) Third you get greater than original. You're adding 50% of the initial value, and decreasing by 33.5% of the new value, which will still give you more than the original

4) Fourth, you get greater than original. You're adding 60% of the initial value, and then decreasing by 40% of the new value, which will still give you greater than the initial value

5) Fifth you get less than the original. You're decreasing by 75%, and then adding 50% of the new value, which still gives you less than the initial value

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A student comes to lecture at a time that is uniformly distributed between 5:09 and 5:14. Independently of the student, the prof
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Answer:

1/2

Step-by-step explanation:

The lecture has already begun when the student arrives means one of these scenarios happen:  

1) the class started at 5:10 and the student arrives at 5:11 or 5:12 or 5:13 or 5:14

2) the class started at 5:11 and the student arrives at 5:12 or 5:13 or 5:14

3) the class started at 5:12 and the student arrives at 5:13 or 5:14

Given student time of arrival is uniformly distributed, then the probability he/she arrives at 5:09 or 5:10 or 5:11 or 5:12 or 5:13 or 5:14 is 1/6.

So,  the probability that the student arrives between 5:11 and 5:14 is 1/6 + 1/6 + 1/6 + 1/6 = 2/3.

The probability that the student arrives between 5:12 and 5:14 is 1/6 + 1/6 + 1/6 = 1/2.

The probability that the student arrives at 5:13 or 5:14  is 1/6 + 1/6 = 1/3.

Given class starting time is uniformly distributed, then the probability it starts at 5:10 or  5:11 or 5:12 is 1/3.

Given the two events are independent, the probability of the first scenario is: (1/3)*(2/3) = 2/9

For the second scenario:  (1/3)*(1/2) = 1/6

For the third scenario:  (1/3)*(1/3) = 1/9

Because all of these scenarios are mutually exclusive the total probability of one of them happen is: 2/9 + 1/6 + 1/9 = 1/2

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