Solve for x: |x| − 8 = −5 (2 points) A. x = −13 and x = −3 B. x = 3 and x = −3 C. x = 3 and x = 13 D. No solution
1 answer:
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Answer:
See attached
Step-by-step explanation:
Table filled in, see picture
- The first column contains points with coordinates (x, y)
- The second column contains the rule (x + (-4), y + 2)
- The third column is obtained by applying the rule. Showing this for one of the vertices:
- Q: (1, 4)
- Apply rule: (1 + (-4), 4 + 2) = (1 - 4, 6) = (-3, 6)
- Get coordinates of Q' from previous step: (-3, 6)
So you get Q'(-3, 6) from Q(-1, 4) by applying the rule (x + (-4), y + 2).
Same steps for other vertices done and shown in the table.
Hope it is more clear.
Answer:
a)
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly less than 4 hours= 240 minutes.
b)
So on this case the 95% confidence interval would be given by (219.345;240.655)
c) H0: µ =240 minutes vs. H1: µ ≠ 240
For this case since the confidence interval contain the value 240. Then we don't have enough evidence to reject the null hypothesis on this case at 5% of significance.
Step-by-step explanation:
Data given and notation
represent the sample mean on minutes
represent the sample standard deviation
sample size
represent the value that we want to test on minutes (because 4*60=240 min)
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
Part a
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 240 minutes, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly less than 4 hours= 240 minutes.
Part b
The confidence interval for the mean is given by the following formula:
(1)
Since the Confidence is 0.95 or 95%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 95% confidence interval would be given by (219.345;240.655)
Part c
H0: µ =240 minutes vs. H1: µ ≠ 240
For this case since the confidence interval contain the value 240. Then we don't have enough evidence to reject the null hypothesis on this case at 5% of significance.