Kyle received the following course grades for 10 (three-credit) courses during his freshman year:
1 answer:
Answer:
(a) The average grade point is 2.5.
(b) The relative frequency table is show below.
(c) The mean of the relative frequency distribution is 0.3333.
Step-by-step explanation:
The given data set is
4, 4, 4, 3, 3, 3, 1, 1, 1, 1
(a)
The average grade point is
Therefore the average grade point is 2.5.
(b)
The relative frequency table is show below:
x f Relative frequency
4 3
3 3
1 4
(c)
Mean of the relative frequency distribution is
Therefore the mean of the relative frequency distribution is 0.3333.
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<h3>Answers:</h3>
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Explanation:
The scale factor is 2, which means we double each coordinate of each point. The general rule is with k = 2. So we can say the more specific dilation rule is
Something like W(-2,2) moves to W ' (-4, 4) after multiplying each coordinate by 2. Do the same for the other points as well.
The given preimage points
- V = (-2, -1)
- W = (-2, 2)
- X = (0, 2)
- Y = (3, -1)
will dilate to the corresponding image points
- V ' = (-4, -2)
- W ' = (-4, 4)
- X ' = (0, 4)
- Y ' = (6, -2)
as shown below. This causes the image to be larger compared to the preimage (since the scale factor is larger than 1). Any given point is twice as far from the origin as compared to before, which in turn means the distance between any two points is twice as much.
Answer:
1) The volume of the frustum is 27·(√3)·h
2) The surface area the frustum is 24·h + 54·√3
Step-by-step explanation:
1) A frustum of the regular triangular prism is a portion of the triangular prism
The length of the edge of the triangle prism = 12
The area of the triangular face, A = (√3/4)·a²
∴ A = (√3/4)·12² = 36·√3
When
The volume of the triangular prism, V = A·h = (36·√3)·h
By triangle proportionality theorem, the cross sectional area of the top half of the triangular prism = (√3/4)·6² = 9·(√3)
The volume of the top half, V₂ = 9·(√3)·h
The volume of the frustum = V - V₂ = (36·√3)·h - 9·(√3)·h = 27·(√3)·h
The volume of the frustum = 27·(√3)·h
2) The surface area of the triangular side of the frustum is given as follows;
A = 36·√3- 9·(√3) = 27·(√3)
The surface area the frustum, = 2×27·(√3) + 2 ×6 ×h + 12×h = 24·h + 54·√3
The surface area the frustum = 24·h + 54·√3