The total cost of 6 wave pool tickets with the discount included would be $73.5.
<h3>How to calculate the total value of 6 tickets with the discount?</h3>
To calculate the total cost we must multiply the value of each ticket by the number of tickets to be purchased, in this case there are 6:
Learn more about tickets in: brainly.com/question/14001767
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Answer:
10 rad 3 / 2
Step-by-step explanation:
x = missing angle of the full traingle
180 - (90 + 60) = x
180 - 150 = x
30 = x
10 rad 3 / 2 = x
this is just my understanding of the problem- im sorry if im incorrect
Using it's concept, the domain of the function is given as follows:
{x| –2 ≤ x <= 5}
<h3>What is the domain of a function?</h3>
The domain of a function is the set that contains all possible input values for the function, that is, the values for which the function is defined.
The function described in this problem is defined from x = -2 until x = 5 hence the domain of the function is given as follows, by the interval:
{x| –2 ≤ x <= 5}
More can be learned about the domain of a function at brainly.com/question/10891721
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Answer:
it is parallel to the y axis
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.