Which properties were used to simplify the following expression? Select all that apply.
2 answers:
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Using the normal distribution, it is found that 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for the amounts are given as follows:
The proportion is the <u>p-value of Z when X = 4250</u>, hence:
Z = 1.66
Z = 1.66 has a p-value of 0.9515.
Hence 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
More can be learned about the normal distribution at brainly.com/question/15181104
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Answer:
.
Step-by-step explanation:
We use the Venn diagram to calculate the desired probabilities.
Note that there are 6 possible results in the sample space
S = {1, 2, 3, 4, 5, 6}
Then note that in the region representing the intercept of A and B there are two possible values.
So
In the region that represents event A there are 4 possible outcomes {4, 5, 1, 2}
So
In the region that represents event B there are 3 possible outcomes {1, 2, 6}
So
.
Now