Answer:
ST = 4
Step-by-step explanation:
Hey there! I'm happy to help!
If 70% of the students are female, then 30% of the students are male. Of these, 85% graduated. Let's find 85% of 30%.
0.85(0.3)=0.255
This means that 25.5% of the students are graduating males.
We want to find the probability that one of the graduating males are picked from the group of graduating people. We have to find how many girls graduated to find the total percent of peole who graduated.
75% of females (70% graduated).
0.75(0.7)=0.525
So, 52.5% of students are graduating females.
We have 52.5% as graduating females and 25.5% are graduating males. We combine this, showing us that 78% of students graduated.
Now, we want to find the probability of picking a graduating male. 25.5% is what percent of 78%? When working with percents, is means equals. Let's say that our percent is p and solve.
0.255=0.78p
We flip the equation so p is on the left.
0.78p=0.255
Divide both sides by 0.78.
p≈0.33 (rounded to nearest hundredth)
Therefore, there is a 33% chance of picking a male from the graduating students.
Have a wonderful day! :D
Step-by-step explanation:
(6x + 7 + x²) + (2x² - 3) =
= 6x + 7 + x² + 2x² - 3 = 3x² + 6x + 4
Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<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.
In this problem, the mean and the standard deviation are given, respectively, by:
The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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