Answer:
The Answer is A.introduction,Body,and ending
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Answer:
3 cars
Step-by-step explanation:
If 12 cars are needed to carry 36 students,
<em>12</em><em>c</em><em>a</em><em>r</em><em>s</em><em> = 36</em>
the number of cars to carry 9 students will be:
xcars = 9
cross the two equations
36x = 12 X 9
36x = 108
x = 108/36
x= 3
Therefore, 3 cars are needed to carry 9 students
OR
12 cars will carry 36 students
so 1 car will carry, (36/12)students
therefore, 1 car will carry 3 students
So, for 9 students,
9students = 9/3 = 3
They all have either a common factor, common multiple, LCF, GCF, LCM, or GCM. The G and L stand for greater and least.
Answer:
- P(x < 84) = 0.3085 or approximately 31%
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<em>Hello to you from Brainly team!</em>
<h3>Given</h3>
- Mean grade μ = 86,
- Standard deviation σ = 4,
- Grade limit x = 84.
<h3>To find </h3>
- Probability of that a randomly selected grade is less than 84 or P(x < 84).
<h3>Solution</h3>
Find z-score using relevant equation:
Substitute values and calculate:
Using the z-score table find the corresponding P- value.
- P(x < 84) = 0.30854 or approximately 31%
Answer:
P ( -1 < Z < 1 ) = 68%
Step-by-step explanation:
Given:-
- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :
u = 67.2
s.d = 4.6
- The random variable (X) that denotes standardized test scores following normal distribution:
X~ N ( 67.2 , 4.6^2 )
Find:-
What percent of the data fell between 62.6 and 71.8?
Solution:-
- We will first compute the Z-value for the given points 62.6 and 71.8:
P ( 62.6 < X < 71.8 )
P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )
P ( -1 < Z < 1 )
- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:
P ( -1 < Z < 1 ) = 68%
P ( -2 < Z < 2 ) = 95%
P ( -3 < Z < 3 ) = 99.7%