Which represents the rotation of ABC toA’B’C a:(x, y) —> (-x,-y)
b:(x, y) —> (x,-y)
c:(x, y) —> (y, -x)

a) A  (-2,-2)  →A¹ (  2, 2)

   B (-5,-2)  →B¹ (5 , 2)

   C (-3,-4) →C¹ ( 3 ,4))

b)

A (-2,-2)  → A¹(  -2, 2)

B (-5,-2)  →B¹ (-5 , 2)

C (-3,-4) →C¹ ( -3 ,4))

c)

A (-2,-2)  → A¹(  -2, 2)

B (-5,-2)  →B¹ (-2 , 5)

C (-3,-4) →C¹ ( -4 ,3))

Step-by-step explanation:

a) Given the (-2,-2) is rotation 180° of counter clock wise then the transformed

to A (-2,-2)  → A¹(  2, 2)

   B (-5,-2)  →B¹ (5 , 2)

   C (-3,-4) →C¹ ( 3 ,4))

b)

Reflection about the x-axis

( x, y) → (  x, -y)

A (-2,-2)  → A¹(  -2, 2)

B (-5,-2)  →B¹ (-5 , 2)

C (-3,-4) →C¹ ( -3 ,4))

c)

( x, y) → (  y, -x)

Given (x, y) is rotation 90° of  clock wise about the origin then the transformed to

A (-2,-2)  → A¹(  -2, 2)

B (-5,-2)  →B¹ (-2 , 5)

C (-3,-4) →C¹ ( -4 ,3))

might be the correct answer will be question number 2

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.