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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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Cos (B) = (a^2 + c^2 -b^2) / (2 * a * c)
cos (B) = (11^2 +17^2 -12^2) / (2 * 11 * 17)
cos (B) = (121 + 289 -144) / (374)
cos (B) = 266 / 374
cos (B) = <span> <span> <span> 0.7112299465
Angle B = </span></span></span>44.665 degrees
cos (B) = (11^2 +17^2 -12^2) / (2 * 11 * 17)
cos (B) = (121 + 289 -144) / (374)
cos (B) = 266 / 374
cos (B) = <span> <span> <span> 0.7112299465
Angle B = </span></span></span>44.665 degrees