Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92
The 90th percentile score is nothing but the x value for which area below x is 90%.
To find 90th percentile we will find find z score such that probability below z is 0.9
P(Z <z) = 0.9
Using excel function to find z score corresponding to probability 0.9 is
z = NORM.S.INV(0.9) = 1.28
z =1.28
Now convert z score into x value using the formula
x = z *σ + μ
x = 1.28 * 92 + 1028
x = 1145.76
The 90th percentile score value is 1145.76
The probability that randomly selected score exceeds 1200 is
P(X > 1200)
Z score corresponding to x=1200 is
z = 
z = 
z = 1.8695 ~ 1.87
P(Z > 1.87 ) = 1 - P(Z < 1.87)
Using z-score table to find probability z < 1.87
P(Z < 1.87) = 0.9693
P(Z > 1.87) = 1 - 0.9693
P(Z > 1.87) = 0.0307
The probability that a randomly selected score exceeds 1200 is 0.0307
Of you have 3 squared you get 9 then 2 times -2 is -4. then you add them 9=-4 which would be ...A: 5
Answer:
First number: 20
Second number: 30
Step-by-step explanation:
First#: x
Second#: y
Constraints:
x + y = 50
1/4x + 2/3y = 25
Isolate variable y:
x + y = 50
-1/4 (x + y = 50 )
-1/4x - 1/4y = -12.5
Add equations the two equations to eliminate x:
-1/4x - 1/4y = -12.5
1/4x + 2/3y = 25
_________________
5/12y = 12.5
5y = 150
y = 30
Substitution to find x:
x + y = 50
x + 30 = 50
x = 20
Answer:
The two triangles are related by A - S - A test of similarity , so the triangles are congruent.