<span>when flipped over a line of reflection the lengths are still the same
the point to the line of reflection is the same length as the line of reflection to the reflected position
the distance from the original point to the reflected point is twice the distance from the original point to the line of reflection. </span>
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
If f(x)=1/9x-2
then f^-1(x) will be
lets assume f(x)=y
y=1/9x-2
y+2=1/9x
9(y+2)=x
f^-1(x)=9(y+2)
rest u can do with the same method
Answer: It should be an 8th grade level
(sorry if i'm wrong)
:(