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grigory [225]
3 years ago
6

A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome. Give the sa

mple space describing all possible outcomes. Then give all of the outcomes for the event of rolling the number 1, 3, or 4. If there is more than one element in the set, separate them with commas. Sample space: {} Event of rolling the number 1 3, or 4 :
Mathematics
1 answer:
Helen [10]3 years ago
7 0

Answer:

Sample space: \Omega=\{1,2,3,4,5,6\}

Event of rolling the number 1 3, or 4 : A={1,3,4}

Step-by-step explanation:

When you roll a number cube with faces labeled from 1 to 6 once.

The possible outcomes are: 1,2,3,4,5 or 6.

Therefore, the sample space of this event is:

  • Sample space: \Omega=\{1,2,3,4,5,6\}

Given the event of rolling the numbers 1, 3, or 4.

Now we are required to give the outcomes for the event of rolling number 1,3 or 4.  Let's call the event A. The set of possible outcomes for A has all the numbers 1, 3 and 4 as follows

  • Event of rolling the number 1 3, or 4 :A= {1,3,4}

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Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)  

t=8.169  

P-value= ?

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a)  P-value=0

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Step-by-step explanation:

From the question we are told that:

Sample size n=10

Mean \=x= 51.3

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Significance level is taken as \alpha=0.05

t test statistics

t=8.169

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P-Value=P(t>8.169)

Critical point

t_{\alpha,df}

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Therefore

P-value from T distribution table

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Conclusion

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We Reject the Null Hypothesis H_0

Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

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