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
#SPJ1
Lets write the problem info into an equation and solve step by step:
7(1/3 + 4/5)
the minimum common multiple of 3 and 5 is 15, so we multiply and divide the fractions by a proper number to convert them to be divided by 15 so is easier to add them:
<span>(1/3)(5/5) = (1*5)/(3*5) = 5/15
(4/5)(3/3) = (4*3)/(5*3) = 12/15
</span>so we substitute in the original equation:
7(1/3 + 4/5<span>)
</span>= 7(5/15 + 12/15<span>)
= 7(17/15)
= (7*17)/15
= 119/15</span>
I do not agree. This is because when the dog was a puppy, she was 370 grams not kilograms. Diego's mistake was mistaking grams for kilograms.
Answer: 18 ft per second
Step-by-step explanation: 36 divided by 2
Answer:
Step-by-step explanation:
C