Y = mx + b....m is the slope and b is the y intercept
slope(m) = 5...(4,0)...x = 4 and y = 0
now we sub our info into y = mx + b...we r looking for b, the y intercept
0 = 5(4) + b
0 = 20 + b
-20 = b
so ur equation is y = 20x - 20...but we need it in standard form
Ax + By = C
y = 20x - 20....subtract 20x from both sides
-20x + y = -20...multiply by -1 to make x positive
20x - y = 20 <== standard form
If you do 50 times 4 it’s easier to find 500 times 400 because all you have to do it add 3 more zeros behind 200 making the answer be 200,000.
-21.85.999
I found the ansewr by dividing both of the numbers to equal tthat
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