Population = 135 students
Mean score = 72.3
Standard deviation of the scores = 6.5
Part (a): Students from 2SD and 3SD above the mean
2SD below and above the mean includes 95% of the population while 3SD includes 99.7% of the population.
95% of population = 0.95*135 ≈ 129 students
99.7% of population = 0.997*135 ≈ 135 students
Therefore, number of students from 2SD to 3SD above and below the bean = 135 - 129 = 6 students.
In this regard, Students between 2SD and 3SD above the mean = 6/2 = 3 students
Part (b): Students who scored between 65.8 and 72.3
The first step is to calculate Z values
That is,
Z = (mean-X)/SD
Z at 65.8 = (72.3-65.8)/6.5 = 1
Z at 72.3 = (72.3-72.3)/6.5 = 0
Second step is to find the percentages at the Z values from Z table.
That is,
Percentage of population at Z(65.8) = 0.8413 = 84.13%
Percentage of population at (Z(72.3) = 0.5 = 50%
Third step is to calculate number of students at each percentage.
That is,
At 84.13%, number of students = 0.8413*135 ≈ 114
At 50%, number of students = 0.5*135 ≈ 68
Therefore, students who scored between 65.8 and 72.3 = 114-68 = 46 students
Answer:
24cb 18 hb
Step-by-step explanation:
Let's call C the number of cheese burger sold and H the number of hamburgers.
As the cheeseburger sold are 3/4 of hamburgers, we can represent with the following equation:
C= (3/4)H (equation 1)
The total of burgers sold are the total of cheeseburger plus the total of hamburgers, and the sum 42. The equation representing it is:
C + H = 42
Replacing C by its value of eq. 1:
(3/4)H + H = 42
(3/4)H + (4/4)H = 42
(7/4) H = 42
Dividing both sides by (7/4)
H = 42/(7/4)
H= 42*(4/7)
H= (42/7)*4
H= 6*4
H=24
Then, C is:
C=(3/4)H
C=(3/4)*24
C= 3*24/4
C=72/4
C=18
We can verify it summing C+H
C+H=24+18=42 verified!
Answer:
0
Step-by-step explanation:
It is not permissible for the denominator to be zero, so the "nonpermissible replacement" for x is 0.
Answer:
The answer is C
Step-by-step explanation:
(180-46)/2=67
67+45=112
Answer:
A&D
Step-by-step explanation:
i think
