Answer:
197,250
Step-by-step explanation:
2.5 * 0.04 * 250 * 8 * 1.25 * 789
0.1 * 250 * 8 * 1.25 * 789
25 * 8 * 1.25 * 789
200 * 1.25 * 789
250 * 789
789/4 * 1000
197.25 * 1000
197,250
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
Consider the attached tree.
You start from the root, and with each level you create a sub-branch for each of the choices you have.
So, the root has two children, white and multigrain bread.
Each of those children has three children, because you have three meat choices
Each of those children has two children, because you have two cheese choices.
Then, you identify all the sandwiches by choosing a leaf, and read the label that lead you there.
For example, leaf number 4 represents a sandwich with white bread, turkey and provolone.