8q/4 hope i helped u out lol
Let X be the score on an english test which is normally distributed with mean of 31.5 and standard deviation of 7.3
μ = 31.5 and σ =7.3
Here we have to find score that separates the top 59% from the bottom 41%
So basically we have to find here x value such that area above it is 59% and below it is 49%
This is same as finding z score such that probability below z score is 0.49 and above probability is 0.59
P(Z < z) = 0.49
Using excel function to find the z score for probability 0.49 we get
z = NORM.S.INV(0.49)
z = -0.025
It means for z score -0.025 area below it is 41% and above it is 59%
Now we will convert this z score into x value using given mean and standard deviation
x = (z* standard deviation) + mean
x = (-0.025 * 7.3) + 31.5
x = 31.6825 ~ 31.68
The score that separates the top 59% from the bottom 41% is 31.68
256×7=1792 in expanded form it is 1000+700+90+2
The nearest hundred thousand is 3,700,000
And the nearest million is 4,000,00
Answer:
- y= 7
- y= -3
- y= -4
- y= -8
Step-by-step explanation:
y= -5x+1
=-5.-1+1
=7
Now you can do this method
