Question 20 of 25
2 answers:
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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:
Least common denominator is 6x^2
Step-by-step explanation:
Here we have denominators x and 6x^2
we take the factors for x and 6x^2
x ---> x
6x^2 ----> 3 * 2* x*x
WE have x in common
To get LCD we multiply all the numbers but repeated x only once
LCD is 3*2*x*x that is 6x^2
Least common denominator is 6x^2