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IceJOKER [234]
3 years ago
12

Help with this algebra question? What goes in the blank?

Mathematics
2 answers:
ANTONII [103]3 years ago
8 0
6x + 3 = x + 13 : Find the value of x by isolating x to one side of the equation.
6x + 3 - x = x + 13 - x
6x + 3 - x = 13 : Subtracting x from both sides removes x from the right side of the equation (because x - x = 0) and maintains the equation's value.
6x + 3 - x - 3 = 13 - 3 : Renoved +3 from the left side of the equation.
6x - x = 10 
5x = 10 
5x / 5 = 10 / 5 : To find the value of x, divide x by 5. You must do the samething on each side of the equation to maintain its value.
x = 2


vlabodo [156]3 years ago
5 0
The answer is rounded to 1.66 because this is actually 1.66 repeated
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The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi
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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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 \mu = 480, \sigma = 20

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.

Z = \frac{X - \mu}{\sigma}

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