We have data for two groups: college juniors represented by the subscript 1 and college seniors represented by the subscript 2. do a two-sample test for equality of means assuming unequal variances. minitab will be of great assistance. when exam-time rolls around, make certain that you are able to calculate tcalc by hand. (a-1) comparison of gpa for randomly chosen college juniors and seniors: x¯1x¯1 = 3.05, s1 = .20, n1 = 15, x¯2x¯2 = 3.25, s2 = .30, n2 = 15, α = .025, left-tailed test.

answ5

step-by-step explanation:

u have to pay atttion to me

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.