Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 10 exams. The first instructor's grades have a variance of 58.9. The second instructor's grades have a variance of 84.8. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%. (Recall: that the larger of the 2 variances needs to be in the numerator as the test is 1-tailed to the right)

Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the  first instructor's variance is smaller and reject H0.

Step-by-step explanation:

1)Formulate the hypothesis that first variance is equal or greater than the second variance

H0: σ₁²≥σ₂²  against the claim  that the first instructor's variance is smaller

 Ha: σ₁²< σ₂²

2) Test Statistic F= s₂²/s₁²

F= 84.8/ 58.9=  1.4397

3)Degrees of Freedom = n1-1= 10-1= 9  and n2 = 10-1= 9

4)Critical value   at 10 % significance level= F(9,9)= 2.4403

5)Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the  first instructor's variance is smaller and reject H0.

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this