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:
If it is, the first and last terms are perfect squares, and the middle term is twice the product of the roots of those squares.
16x^ = (4x)^2 . . . . a perfect square
49 = 7^2 . . . . . . . a perfect square
-2(4x)(7) = -56x . . . . . matches the middle term, so the factoring is ...
16x^2 -56x +49 = (4x -7)^2
_____
The sign in the binomial factors is the same as the sign of the middle term of the trinomial.
Answer: 321 adult tickets and 227 children tickets were sold.
Step-by-step explanation:
Let x represent the number of adult tickets that were sold.
Let y represent the number of children tickets that were sold.
The total number of tickets that the theatre sold is 548. This means that
x + y = 548
Adult tickets sell for $6.50 each, and children's tickets sell for $3.50 each. The total ticket sales was $2881. This means that
6.5x + 3.5y = 2881 - - - - - - - - - - -1
Substituting x = 548 - y into equation 1, it becomes
6.5(548 - y) + 3.5y = 2881
3562 - 6.5y + 3.5y = 2881
- 6.5y + 3.5y = 2881 - 3562
- 3y = - 681
y = - 681/ -3
y = 227
x = 548 - y = 548 - 227
x = 321
23 Pennie’s is the answer I believe
Answer
150x-2x=3
Step-by-step explanation:
