36 kg bag of cement was dropped losing one tenth of cement. How much was left in the bag of cement ?
2 answers:
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Answer:
<h2>Step-by-step explanation:
-5(m - 3) + 6 = 2(m - 5)
<u>Expand the terms in the bracket</u>
That's
- 5m + 15 + 6 = 2m - 10
- 5m + 21 = 2m - 10
<u>Subtract 2m from both sides</u>
- 5m - 2m + 21 = 2m - 2m - 10
- 7m + 21 = - 10
<u>Subtract 21 from both sides</u>
That's
- 7m + 21 - 21 = - 10 - 31
- 7m = - 31
<u>Divide both sides by - 7</u>
We have the final answer as
<h3>Hope this helps you
Answer:
A customer who sends 78 messages per day would be at 99.38th percentile.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Average of 48 texts per day with a standard deviation of 12.
This means that
a. A customer who sends 78 messages per day would correspond to what percentile?
The percentile is the p-value of Z when X = 78. So
has a p-value of 0.9938.
0.9938*100% = 99.38%.
A customer who sends 78 messages per day would be at 99.38th percentile.
