A hyperbola centered at the origin has a vertex at (-6, 0) and a focus at (10, ).

Vertices: (0,-a),(0,a)

Foci: (0,-c).(0.c)

Asymptotes: y==

Vertices: (-a,0),(a,0)

Foci: (-0,0),(0,0)

Asymptotes your

Directrices --

Standard Equation

Directrices

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Standard Equation

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Which are the equations of the directrices?

answer: i think you should do the math and pay attention in class : )

step-by-step explanation:

that is truly what   think but if i get the answer i will

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.