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horsena [70]
3 years ago
8

A student earned grades of 78, 72, 74, and 88 on her four regular tests. She earned a grade of 79 on the final exam and 88 on he

r class projects. Her combined homework grade was 89. The four regular
tests count for 40% of the final grade, the final exam counts for 20%, the project counts for 10%, and homework counts for 30% What is her weighted mean grade? Round to one decimal place.
O A. 82.5
OB. 71.0
O C. 63.6
OD. 81.5
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
3 0

Answer:

A: 82.5

Step-by-step explanation:

Average score of the 4 regular tests = (78 + 72 + 74 + 88)/4 = 78

We are told;

- the regular tests account for 40% of the final grade.

- final exam counts for 20%

- project counts for 10%

- homework counts for 30%

Thus;

weighted mean grade = (40% × 78) + (20% × 79) + (10% × 88) + (30% × 89) = 82.5

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3 years ago
The water level of a certain reservoir is depleted at a constant rate of 1000 units daily. The reservoir is refilled by randomly
Aloiza [94]

Answer:

The probability that the reservoir will be empty in the next ten days is 0.476

Step-by-step explanation:

In 10 days, the reservoir will have 10000 units of water depleted. Since the reservoir starts with almost 5000, then by gaining 8000 units it wont get emptied. It also wont get emptied if it gains 5000 or more units twice, thus, if it rains during 2 days it shoudnt get emptied.

However, note that the reservoir could get emptied if it doesnt rain in the first five days, because it will lose 5000 units, exceding its starting amount. Thus, there are only 2 possibilities for the reservoir to be emptied:

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  • If it rains only once in the fisrt five days providing only 5000 units of water, and during the second period of five days it doesnt rain.

We know that the rainfall distribution is Poisson with rate 0.2 per day. In five days the rate is 0.2*5 = 1; lets call X this random variable. The probability to not have any rainfalls in 5 days is

P_X(0) = \frac{e^{-1} 1^0}{0!} = e^{-1} = 1/e

Also, the probability for it to rain just once in five days is the probability of X being equal to one

P_X(1) = \frac{e^{-1} 1^1}{1!} = 1/e

Therefore, the probability of the event 'it doesnt rain in the first five days' is 1/e, and the probability of the other event 'it rains once in the first five days, also that rain adds 5000 units of water and it doesnt rain in the other five days' is

1/e*0.8*1/e  (we multiply 1/e foe the probability that the rain gives 5000 units of water and then by the probability that it doesnt rain in the last five days).

Summing both disjoint events, we obtain that the probability that the reservoir is emptied is

1/e+ 1/e*0.8*1/e = 0.476

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Step-by-step explanation

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