At Monroe High School, 62% of all students participate in after school sports and 11% participate in both after school sports an
2 answers:
You might be interested in
Chebyshev's theorem in statistics states that for many probability distributions, no more than 1/k² of measured values will be k standard deviations away from the mean.
Because the area under the probability distribution curve is equal to 1, Chebyshev's theorem means that the shaded area shown in the figure is equal to 1 - 1/k².
When k = 1.75, the shaded area is
1 - 1/1.75² = 0.7635 = 67.35%
Therefore the percent of the area within +/- 1.75 standard deviations from the mean is
67.35/2 = 33.7%, which is at least 33% of the observations.
Answer:
According to the Chebyshev theorem, at least 33% of the observations lie within +/- standard deviations from the mean.
Because the area under the probability distribution curve is equal to 1, Chebyshev's theorem means that the shaded area shown in the figure is equal to 1 - 1/k².
When k = 1.75, the shaded area is
1 - 1/1.75² = 0.7635 = 67.35%
Therefore the percent of the area within +/- 1.75 standard deviations from the mean is
67.35/2 = 33.7%, which is at least 33% of the observations.
Answer:
According to the Chebyshev theorem, at least 33% of the observations lie within +/- standard deviations from the mean.
Given:
Vertices of ABCD are A(-4,4), (-2,2), C(-2,-1) and D(-4,1).
Vertices of A'B'C'D' are A'(3,-4), B'(5,-2), C'(5,1) and D'(3,-1).
To find:
The sequence of transformations that changes figure ABCD to figure A'B'C'D'.
Solution:
Part A:
The figure ABCD reflected across the x-axis, then
Using this rule, we get
Similarly, the other points are .
Then figure translated 7 units right to get A'B'C'D'.
Similarly, the other points are .
Therefore, the figure ABCD reflected across the x-axis and then translated 7 units right to get A'B'C'D'.
Part B:
Reflection and translation are rigid transformation, it means shape and size of figures remains same after reflection and translation.
Therefore, the two figures congruent.
