9) Final Exam scores in a math class are Normally distributed with a mean of 75. A faculty member believes that the mean is less than 75 and decides to test his claim at a 5% significance level by taking a random sample of Final Exam scores from 7 students. The 7 scores were: 55 61 94 50 71 64 81 Determine: Но H p-value p = Decision: Reject H OR 0 Do not reject H 0 Conclusion:

Q: 43=31-5(1-5w)

A: Given that,   43 = 31 - 5(1-5w) We have to solve for w.

Q: 8 15

A:

Q: 3y² - 2y-1 (-)-5y²- 2y +6

A:

Q: 3. 42P 4-2p-1 =

A:

Q: Todt 5(-37+4)

A: Given,   ∫  -3t+410dt

Q: 2'dr

A: Explanation and solution is given below....

Q: 5y + 6(1-3)=15

A: Answer y = 3  solution is given in step 2

Q: 2n +2 = n=1 O 28 O 14 O 10

A:

Q: d. 1 3a+18 = 1 - 8 a+6

A:

Q: 4(P-9) = 92

A: Answer

Q: n=5 II

A:

Q: dy 3 メ 2

A: SINCE YOU HAVE ASKED MULTIPLE QUESTIONS IN SINGLE REQUEST, WE WILL BE ANSWERING ONLY THE FIRST…

Q: 43a+10-26a=27

A: To solve: 43a+10-26a=27

Q: -2 -4 and D = 4 8

A:

Q: с 4 B 12 A

A:

Q: 12  6 7    −    2    =

A: To find the value of 1267-2

Q: onn -||-|

A: