Answer:
Step-by-step explanation:
We are given that a student needs to make a square cardboard piece.
Perimeter of cardboard should be equal to atleast 92 inches
We have to find that which shows reasonable domain for f(s)
Let s be the side of square cardboard
We know that perimeter of square =f(s=)
Then, perimeter of cardboard=
Dividing by 4 on both sides
Hence, the domain of function
Therefore, option d is true.
Answer:d:
Answer:
Step-by-step explanation:
1) Data given and notation
n=100 represent the random sample taken
X=18 represent the ounce cups of coffee that were underfilled
estimated proportion of ounce cups of coffee that were underfilled
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed for this case is . The next step would be calculate the p value for this test.
Since is a one right tailed test the p value would be:
So the p value obtained was a very low value and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .