Answer:
Step-by-step explanation:
Corresponding scores before and after taking the course form matched pairs.
The data for the test are the differences between the scores before and after taking the course.
μd = scores before taking the course minus scores before taking the course.
a) For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
b) We would assume a significance level of 0.05. The P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.
Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.
c) The mean difference for the sample scores is greater than or equal to zero
Divide all the ratios and compare:
8/12 = 0.666
15/10 = 1.5
2/3 = 0.666
6/9 = 0.666
The one that isn't equivalent is 15 over 10
Step-by-step explanation:
29+1/5r=
29+(1/5×-1/2)
29+1/10
291/10
=29.1
Answer: v=42
Step-by-step explanation:
18+(6)(4)
Answer:

Step-by-step explanation:



