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iren2701
2 months ago
11

A set of face cards contains 4 Jacks, 4 Queens, and 4 Kings. Carlie chooses a card from the set, records the type of card, and t

hen replaces the card. She repeats this procedure a total of 60 times. Her results are shown in the table.
How does the experimental probability of choosing a Queen compare with the theoretical probability of choosing a Queen?

The experimental probability is 4 less than the theoretical probability.
The experimental probability is 1/15 less than the theoretical probability.
The experimental probability is 1/15 more than the theoretical probability.
The experimental probability is 4 more than the theoretical probability.

Mathematics
2 answers:
Leona [12.6K]2 months ago
8 0

Answer:

Option b is correct

PIT_PIT [12.4K]2 months ago
8 0
<span>The accurate answer is "<span>The experimental probability is 1/15 lower than the theoretical probability."

The attached table illustrated in the image indicates that the theoretical probabilities are: 2/5 for drawing a Jack, 4/15 for a Queen, and 1/3 for a King, based on the frequency of each card in the set.
</span>
</span>
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Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
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Answer:

Step-by-step explanation:

It has been established that the count of drivers traveling between a specific origin and destination in a certain time frame follows a Poisson distribution with a mean μ = 20 (as indicated in the article "Dynamic Ride Sharing: Theory and Practice"†).

a) P(X\leq 15) = 0.1565=0.157

b) P(X>26) =1-F(26)\\= 1-0.9221\\=0.0779=0.078

c) P(15\leq x\leq 26)\\=F(26)-F(14)\\=0.9221-0.1049\\=0.8172=0.817

d) 2 standard deviations = 2(20) = 40

Thus, this means the range for 2 standard deviations is

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1 month ago
Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B
Zina [12379]

Response:

a. 0.76

b. 0.23

c. 0.5

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p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed explanation:

a. p(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

        = 0.6 + 0.4 + 0.2 - 0.3 - 0.11 - 0.1 + 0.07 = 0.76

b. P(AnBnC') = P(AnB) - P(AnBnC)

        = 0.3 - 0.07 = 0.23

c. P(B/A) = P(AnB)/P(A)

        = 0.3/0.6 = 0.5

e. P((AnB)/C) = P((AnB)nC)/P(C)

        = P(AnBnC)/P(C)

        = 0.07/0.2 = 0.35

f. P((AUB)/C) = P((AUB)nC)/P(C)

        = (P(AnC) U P(BnC))/P(C)

        = (0.11 + 0.1)/0.2

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7 0
1 month ago
How to solve 67.8 × 0.45
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26 days ago
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Answer:

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Step-by-step explanation:

A scatter plot illustrates the relationships between two variables for an individual. It is essentially a graph in which a best-fit curve is drawn to encapsulate the complete dataset. A scatter plot is considered to have a robust correlation if the correlation coefficient is near r = 1, indicating a very strong connection between the two variables.

A scatter plot exhibits a solid correlation when its data points are closely aligned to the line or curve of best fit.

For r = 1, the correlation is regarded as strong and positive.

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Because

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To convert 100 USD: 100 × 113.83 Japanese Yen

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