Answer: The answer is D .
Step-by-step explanation:
6m² + 6m - 9m²
Add like terms.
(6m² - 9m²) + 6m
-3m² + 6m
~Hope I helped!~
Answer:
-186
Step-by-step explanation:
When you add a positive number to a negative, it gets bigger (In value) . For example:
-5 + 4 = -1
-10 + 12 = 2
So for this equation, -200 + 14, all you have to do is subtract 14 from 200, which will get you 186.
Since you removed the negative, add it back to get -186
<em>they </em><em>both </em><em>would </em><em>have </em><em>2</em><em>4</em>
<em> </em><em>they </em><em>both </em><em>would </em><em>have </em><em>2</em><em> </em><em>successful</em><em> </em><em>shots.</em><em> </em><em>they </em><em>would </em><em>mean </em><em>that </em><em>the </em><em>older </em><em>brother</em><em> </em><em>added </em><em>4</em><em> </em><em>point </em><em>to </em><em>his </em><em>score </em><em>and </em><em>the </em><em>younger </em><em>brother</em><em> </em><em>would </em><em>have </em><em>added </em><em>6</em><em> </em><em>to </em><em>his </em><em>score</em>
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