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docker41 [41]
3 years ago
7

Tyson has two movie tickets and wants to randomly select one of six friends to go with him. He rolls a six-sided die to make his

decision. He tested this model 60 times and recorded the results in this table.
Result of Roll Times Result Occurred
1 - 11
1 - 0.18
2 - 9
2 - 0.15
3 - 9
3 - 0.15
4 - 10
4 - 0.17
5 - 11
5 - 0.18
6 - 10
6 - 0.17
Do you think the relative frequency from the experiment is a good predictor of the theoretical probability? Why or why not?
Mathematics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

<u><em>The relative frequency of rolling a particular number can be calculated using the formula </em></u>

<u><em> </em></u>

<u><em>relative frequency , where f is the actual frequency of an event and n is the number of times the experiment is performed. This experiment had the following results: </em></u>

<u><em> </em></u>

<u><em>The relative frequency of rolling a 1 is 0.2. </em></u>

<u><em>The relative frequency of rolling a 2 is about 0.23. </em></u>

<u><em>The relative frequency of rolling a 3 is about 0.13. </em></u>

<u><em>The relative frequency of rolling a 4 is 0.15. </em></u>

<u><em>The relative frequency of rolling a 5 is 0.15. </em></u>

<u><em>The relative frequency of rolling a 6 is about 0.13. </em></u>

<u><em>The relative frequencies of rolling 1, 2, 3, 4, 5, and 6 are quite similar. So, the relative frequency is a good predictor of the theoretical probability. </em></u>

Step-by-step explanation:

this is exact answer from edmentum so change it up a bit

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Answer:

Minimum unit cost = 5,858

Step-by-step explanation:

Given the function : C(x)=x^2−520x+73458

To find the minimum unit cost :

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dC/dx = 2x - 520

Set = 0

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2x = 520

x = 260

To minimize unit cost, 260 engines must be produced

Hence, minimum unit cost will be :

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Sonbull [250]

Answer:

μ = 235.38

σ = 234.54

Step-by-step explanation:

Assuming the table is as follows:

\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]

This is an example of grouped data, where a range of values is given rather than a single data point.  First, find the total frequency.

n = 339 + 86 + 55 + 18 + 11 + 8 + 3

n = 520

The mean is the expected value using the midpoints of each range.

μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520

μ = 122400 / 520

μ = 235.38

The variance is:

σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)

σ² = 55009.7

The standard deviation is:

σ = 234.54

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3 years ago
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Entry ticket = $7

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Inequality => 7+g ≤ 30

Hope it helps!

6 0
2 years ago
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