Music students often don’t evaluate their own performances accurately. can small-group discussions ?
a) the subjects were 29 students preparing for the end-of-semester performance that is an important part of their grade.
b) the 15 students in one group each videotaped a practice performance, evaluated it themselves, and then discussed the tape with a small group of other students.
c) the remaining 14 students watched and evaluated their tapes alone.
d) at the end of the semester, the discussion-group students evaluated their final performance more accurately

4.08%

step-by-step explanation:

14

step-by-step explanation:

14 x 3 = 42

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.