Suppose 10,000 students have taken a certain course and all of them were invited to evaluate the course. Out of the 10,000, we see that 496 responded to the course evaluation. And out of those 496, there were 130 that gave the course the highest rating (that is, 5 on a scale from 1 to 5). We want an approximate confidence interval for p, the proportion of all 10,000 students that would give this course the highest rating. We estimate p to be 130/496 or about 26.2%. Is a z-interval for a proportion applicable here

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.

your mom

step-by-step explanation: