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matrenka [14]
3 years ago
11

50,000 contestants participate in an on-line game. The game randomly eliminates 20% of the contestants each day. Determine the n

umber of contestants remaining in the game after one week.
Mathematics
1 answer:
irga5000 [103]3 years ago
6 0

Answer:10,486 contestants

Step-by-step explanation:

First day:

Total number of contestants at the beginning = 50,000

After first day, 20% 0f 50,000 are removed.

That is, (20/100)×50,000 = 10,000

Contestants remaining after first day = 50,000-10,000 = 40,000

Second day:

Contestants= 40,000

20% removed

That is (20/100)×40000=8000

Balance =40000-8000=32,000

Third day:

Contestants= 32000

20% removed

That is (20/100)×32000=6400

Balance =32000-6400=25,600

Fouth day:

Contestants= 25,600

20% removed

That is (20/100)×25,600=5,120

Balance =25,600-5,120=20,480

Fifth day:

Contestants= 20,480

20% removed

That is (20/100)×20,480=4096

Balance =20,480-4096=16,384

Sixth day:

Contestants= 16,384

20% removed

That is (20/100)×16,384=3276.8

Balance =16,384-3276.8=13,107.2

Seventh day:

Contestants= 13,107.2

20% removed

That is (20/100)×13,107.2=2621.44

Balance =13107.2-2621.44=10485.76

Therefore, number of contestants remaining after one week equals 10,486.

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Step-by-step explanation:

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Solution to the problem

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

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3 years ago
I got 1.1... is that right?
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No, the answer is 2

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