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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Solve for r in the first equation:
3(r+300) = 6
Use the distributive property:
3r + 900 = 6
Subtract 900 from both sides:
3r = -894
Divide both sides by 3:
r = -894 / 3
r = -298
Now you have r, replace r in the second equation and solve:
r +300 -2 =
-298 + 300 - 2 = 0
The answer is 0.
Answer:
y=20x + 40x
Step-by-step explanation:
Answer:
40
Step-by-step explanation:
Let the number be y
-1/2 × y = -20
y = -20 ÷ -1/2
y = 20 × 2
y = 40