Answer:
$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
Normally distributed with mean $480 and standard deviation $20.
This means that 
How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05?
This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.




$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05
Answer:
B & E
Step-by-step explanation:
___________________
-6x + 12
To find this answer, follow these steps:
First, multiply 2x + 3 and x - 6 using either the FOIL method or the box method. This should get you the answer of 2x^2 - 9x - 18
This should now replace the 2x + 3 and x - 6
So with that, you should have this currently:
2x^2 - 9x - 18 - 2x^2 + 3x + 30
Now, combine like terms:
The 2x^2 cancels each other out. After that, combine -9x and 3x to get -6x. Then -18 and 30.
This should leave you with the final answer of -6x + 12
Hope this helps!
I believe the answer is about a 60% chance. I hope this helps!