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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12⁷/₁₂ - 5¹⁴/₅
12⁷/₁₂ - 7⁴/₅
12³⁵/₆₀ - 7⁴⁸/₆₀
11⁹⁵/₆₀ - 7⁴⁸/₆₀
4⁴⁷/₆₀
Answer:
C. 1:27
Step-by-step explanation:
a²/b²=1/9 => a=1 and b=3
a³/b³=1³/3³=1/27
C. 1:27
$500 < or = $415 + d
(subtract the 415 from both sides)
$85 < or = d
Answer:
+3.8 to each side
Step-by-step explanation: