Acamera is placed in front of a hyperbolic mirror. the equation of the hyperbola that models the mirror is y^2/16-x^2/9=1 where x and y are in inches the camera is pointed toward the vertex of the hyperbolic mirror and is positioned such that the lens is at the nearest focus to that vertex. the lens inch(es) from the mirror.

The lens is 6 inches from the mirror

Step-by-step explanation:

We need to find the distance of the lens from the mirror, if  the camera is pointed toward the vertex of the hyperbolic mirror and is positioned such that the lens is at the nearest focus to that vertex.

Given:

where a² = 16 and b² = 9

Then c² = a² + b²

c² = 16+9

c² = 25

c= √25 = 5

The hyperbola is centered at (0,0)

The vertex is at (0,a) = (0,4)

The foci is at (0,c) = (0,5)

The distance from vertex to focus is

d= c-a = 5-4 = 1 inches

Total distance from lens to mirror is sum of distance d and foci c:

5+1= 6 inches

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

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.