The money collected from selling bacon at a butcher store is given by the function f(x) = 3.55x – 4, where f(x) is the sales rev
2 answers:
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Step-by-step explanation:
- First, the radical needs to be alone on one side of the equation, if there is - 4, - 3 next to it then you'll have to transfer it to the opposite side like that:
5. (y+7)^0.5 - 4=4
(y+7)^0.5=8......
- Second, you need to square the two sides of the equation :
5.((y+7)^0.5) ^2=8^2
Y+7=64
Y+7-7=64-7
Y=57
Note: if the radical's index is (2) then you'll square the two sides of the equation, but if the index is (3) you'll have to cube the two sides of the equation, "look at the exercises 9,10"
Answer:
A. t ≥ 1.734
Step-by-step explanation:
Data given and notation
represent the mean for the sample male
represent the mean for the sample female
represent the sample standard deviation for the males
represent the sample standard deviation for the females
sample size for the group male
sample size for the group female
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the mean for females is higher than the mean for males, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:
(1)
And the degrees of freedom are given by
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
What is the test statistic?
With the info given we can replace in formula (1) like this:
Critical value
For this case since we have a right tailed test we need to look into the t distribution with 18 degrees of freedom a value that accumulates 0.05 of the area on the right, and on this case:
And the rejection zone of the null hypothesis would be: A. t ≥ 1.734
For our case our calculated value is higher than the critical value so we have enough evidence to reject the null hypothesis at the 5% of significance.