In a sample of n = 6 scores, 5 of the scores are each above the mean by one point. Where is the 6th score located relative to the mean?

answer: 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

step-by-step explanation:

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.

one triangle is 2.5 times bigger than the other triangle.

the given growth  function for the length of the fish (cm) is

f(m) = 5(1.07)m = 5.35m

where 

m = months

at the end of the study, the length is 9.19 cm. the  time for conducting the study is determined as

5.35m = 9.19

m = 1.7178 months

part a.

the study was performed over a time period  of about 1.72 months. therefore  the domain for  plotting the function is 0 < = m < = 1.72.

note:

if growth  continues at this rate for up to m months, then the domain would be 0< =m< =m.

part b.

the growth function is a linear function.

when m=0, f=0.

therefore the initial growth is zero, which represents the y-intercept.

part c.

assume that the growth function is valid for 0< =m< =9.

the average rate of change of the function is the slope of the linear function, which is 5.35 cm/month.