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Travka
2 months ago
7

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

ental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?
Mathematics
2 answers:
babunello [11.8K]2 months ago
7 0

Answer:

Juan's assertion is inaccurate. The actual experimental probability is StartFraction 2 over 9 EndFraction.

Step-by-step explanation:

took the test.

babunello [11.8K]2 months ago
3 0

Answer:

\frac{2}{9}

Step-by-step explanation:

Given:

Juan rolled a die with six faces 18 times.

The outcome of two was recorded four times.

To Find: Juan estimated the experimental likelihood of landing on two to be approximately 1/9. Why is his probability estimation wrong?

Solution:

Total rolls = count of rolls = 18

Favorable rolls = The result of two happened four times.  = 4

Thus, the experimental probability = \frac{\text{Favorable events}}{\text{Total events}}

                                               = \frac{4}{18}

                                               = \frac{2}{9}

Therefore, the experimental probability of rolling a two is \frac{2}{9}

This indicates that Juan's experimental probability was incorrect.

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There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
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2 months ago
According to Consumer Digest (July/August 1996), the probable location of personal computers (PC) in the home is as follows: Adu
PIT_PIT [12445]

Answer:

a) 0.32

b) 0.68

c) Office or den

Step-by-step explanation:

a) The likelihood that a personal computer is located in a bedroom can be calculated by adding the probabilities of it being in an adult bedroom, child bedroom, or another bedroom:

P(B) =P(adult)+P(child)+P(other)\\P(B) = 0.03+0.15+0.14\\P(B) =0.32

b) The probability of a PC not being located in a bedroom is found by calculating 100% minus the probability of it being in a bedroom:

P(Not\ B) = 1- P(B)\\P(Not\ B) =1-0.32\\P(Not\ B) =0.68

c) The expected location for finding a personal computer in a randomly selected household is derived from the room most likely to contain a PC according to Consumer Digest. The Office or den ranks as the most likely place with a probability of 0.40.

A PC would most likely be found in the Office or den.

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2 months ago
Sal recorded the number of crackers in each snack-sized package he opened. Which statement must be true according to the box plo
PIT_PIT [12445]

Response:

The data shows skewness, with the minimum amount of crackers in a pack being 7

Detailed explanation:

Hello,

Firstly, the question lacks completeness due to missing information from the box plot, which I have provided to assist you in answering your inquiry.

Considering the details from the attached image, a symmetric distribution would be centered evenly, but that is not the case here.

The image indicates a positive skew, with the lowest count recorded as 7.

7 0
1 month ago
Read 2 more answers
Jack and jill exercise in a 25.0-m-long swimming pool. jack swims nine lengths of the pool in 2 minutes and 34.5 seconds, while
lawyer [12517]
Jack swims a total distance of 9*25.0m=225.0 meters in 2 minutes and 34.5 seconds.

The time duration of 2 minutes and 34.5 seconds can be converted to seconds: 2*60sec +34.5 sec= 154.5 s


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\displaystyle{ Average\ Speed= \frac{Distance\ traveled}{Time \ of \ travel}\\\\



\displaystyle{ Average \ Velocity = \frac{Displacement}{time}



Therefore,
the speed of Jack calculates to: 225.0 meters / 154.5 s =(225/154.5) m/s = 1.46 m/s

the speed of Jill calculates to: 250.0 meters / 154.5 s =(250/154.5) m/s = 1.62 m/s


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Thus, Jill's average Velocity is 0,

and Jack's average velocity is 25/154.5 = 0.16 m/s

Answer:

Jack's Average Speed: 1.46 m/s

Jill's Average Speed: 1.62 m/s

Jack's Average Velocity: 0

Jill's Average Velocity: 0.16 m/s

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1 month ago
Doug purchased land for $8,000 in 1995. The year value of the land depreciated by 4% each year thereafter. Use an exponential fu
Leona [12618]

Answer:

Choice C. $6,012

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We know that

The formula to find the depreciated value is given by

V=P(1-r)^{t}

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P represents the original value

r corresponds to the depreciation rate in decimal

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r = 0.04

Plug these into the formula mentioned earlier

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Hope this assists you:)

4 0
2 months ago
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