Answer:


a. What is the probability that the cost will be more than $450?
We are supposed to find P(x > 450)
Formula : 


Refer the z table :
P(z<0.9431)=0.8264
P(z> 0.9431)=1-P(z<0.9431)=1- 0.8264= 0.1736
Hence the probability that the cost will be more than $450 is 0.1736
b)What is the probability that the cost will be less than $250?
We are supposed to find P(x<250)
Formula : 


Refer the z table :
P(z<-1.3295)=0.0934
Hence the probability that the cost will be less than $250 is 0.0934
c)What is the probability that the cost will be between $250 and $450?
P(250<z<450)=P(z<450)-P(z<250) = 0.8264 - 0.0934 =0.733
Hence the probability that the cost will be between $250 and $450 is 0.733
d) If the cost for your car repair is in the lower 5% of automobile repair charges, what is your cost?
p = 0.05
refer the z table
z = -1.65
Formula : 




Hence If the cost for your car repair is in the lower 5% of automobile repair charges, so, cost is $221.8
Answer:
0
Step-by-step explanation:
We have the fraction
Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:
![\frac{(2m-3)(2m)-(2m+3)(2m)}{(2m+3)(2m-3)} =\frac{2m^2-6m-[2m^2+6m]}{(2m+3)(2m-3)} =\frac{2m^2-6m-2m^2-6m}{(2m+3)(2m-3)} =\frac{-12m}{(2m+3)(2m-3)}](https://tex.z-dn.net/?f=%5Cfrac%7B%282m-3%29%282m%29-%282m%2B3%29%282m%29%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B2m%5E2-6m-%5B2m%5E2%2B6m%5D%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B2m%5E2-6m-2m%5E2-6m%7D%7B%282m%2B3%29%282m-3%29%7D%20%3D%5Cfrac%7B-12m%7D%7B%282m%2B3%29%282m-3%29%7D)
Step 2. Equate the resulting fraction to zero and solve for
:

![-12m=0[(2m+3)(2m-3)]](https://tex.z-dn.net/?f=-12m%3D0%5B%282m%2B3%29%282m-3%29%5D)



Step 3. Replace the value in the original equation and check if it holds:


Since
,


Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution
Answer:
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?
This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So
X = 12



has a pvalue of 0.4052
X = 8



has a pvalue of 0.0329
0.4052 - 0.0329 = 0.3723
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade