In the spring of 2016, he develops on-line videos as a supplement to his class lecture. During that semester, his average grade is 71% from a sample of 49 students. The level of significance is 2.5%. What is the correct conclusion?a. Reject the null hypothesis, conclude the grades improvedb. Fail to reject the null hypothesis, conclude the grades did not improve
c. Reject the null hypothesis, conclude the grades did not improve
d. Fail to reject the null hypothesis, conclude the grades did improve

(c) Reject the null hypothesis, conclude the grades did not improve.

Step-by-step explanation:

Null hypothesis: The grades improved.

Alternate hypothesis: The grades did not improve.

Test statistic (z) = p ÷ sqrt[p(1-p) ÷ n]

p is sample proportion = 0.71

n is sample size = 49 students

z = 0.71 ÷ sqrt[0.71(1-071) ÷ 49] = 0.71 ÷ 0.065 = 10.92

The test is a two-tailed test. At 2.5% significance level, the critical values are -1.96 and 1.96.

Conclusion:

Reject the null hypothesis because the test statistic 10.92 falls outside the region bounded by the critical values -1.96 and 1.96.

Since the null hypothesis is rejected, it therefore means the grades did not improve.

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: