Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<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 are given, respectively, by:

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:


Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:


Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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if there are 300 raffle tickets and one raffle ticket is drawn, the proabability of drawing the next one is 1 in 299 because there is one less ticket from the previous win
Answer:
option B
Given : |x + 4| < 5
A. –5 > x + 4 < 5
B. –5 < x + 4 < 5
C. x + 4 < 5 and x + 4 < –5
D. x + 4 < 5 or x + 4 < –5
In general , |x|< n where n is positive
Then we translate to -n < x < n
|x + 4| < 5
5 is positive, so we translate the given absolute inequality to
-5 < x+4 < 5
So option B is correct
3 1/2r=28
3.5r=28
cross of 3.5 and r remains
divide 28 and 3.5
and r=8
r=8
The GCF of 6^3, -9^2, and 12x is 3x, so you could rewrite it as
3x(2x^2-3x+4), hope this helps!