Since the histogram is not symmetric, the grades shown in the math class below are not normally distributed.
<h3>When does a histogram represent a normal distribution?</h3>
A histogram represents a normal distribution if it symmetric.
In this problem, we have that:
- 57% of the grades are on the left tail.
- 25% of the grades are on the center.
- 18% are on the right tail.
Since the percentages at the tails are different, the histogram is not symmetric, and the grades shown in the math class below are not normally distributed.
More can be learned about the normal distribution at brainly.com/question/24537145
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The value of

,

, and

for each schools is given in the first picture below.

is the lower quartile

is the median

is the upper quartile
School A:
Minimum value is 2
Maximum value is 22
The lower quartile is 2.5
The median is 10
The upper quartile is 15.5
School B:
Minimum value is 9
Maximum value is 20
The lower quartile is 12
The median is 16
The upper quartile is 18
The box plot for each school is shown in the second picture
Box plot for school A isn't symmetrical. The data tails on the right
Box plot for school B isn't symmetrical. The data tails on the left
1.
Answer A. (x<1)
2.
Answer A.
3.
Answer A. (x<0)
4.
Answer D. (x>2.5)
Answer:
44/15 or maybe 2 14/15
Step-by-step explanation:
Answer:
C. x = 3, y = 2
Step-by-step explanation:
If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.
Thus:
x + 3 = 3y (eqn. 1) => equal hypotenuse
Also,
x = y + 1 (eqn. 2) => equal legs
✔️Substitute x = y + 1 into eqn. 1 to find y.
x + 3 = 3y (eqn. 1)
(y + 1) + 3 = 3y
y + 1 + 3 = 3y
y + 4 = 3y
y + 4 - y = 3y - y
4 = 2y
Divide both sides by 2
4/2 = 2y/2
2 = y
y = 2
✔️ Substitute y = 2 into eqn. 2 to find x.
x = y + 1 (eqn. 2)
x = 2 + 1
x = 3