Quadrilateral A’B’C’D’ is a dilation of quadrilateral ABCD about point P. Quadrilateral ABCD is shown. Side AB is labeled 5. Sid
e BC is labeled 4. Side CD is labeled 10. Side DA is labeled 6. Quadrilateral A prime B prime C prime D prime is between quadrilateral ABCD and point P. Side A prime B prime is labeled 1 and one fourth. Side B prime C prime is labeled one. Side C prime D prime is labeled 2 and a half. Side D prime A prime is labeled 1 and a half. reduction enlargement
2 answers:
Answer: The correct option is (A) reduction.
Step-by-step explanation: Given that the quadrilateral A'B'C'D' is a dilation of the quadrilateral ABCD.
As shown in the given figure, the lengths of the sides of quadrilateral ABCD are as follows:
AB = 5 units, BC = 4 units, CD = 10 units and DA = 6 units.
And, the lengths of the sides of quadrilateral A'B'C'D' are as follows:
We know that the dilation will be an enlargement if the scale factor is greater than 1 and it will be a reduction if the scale factor is less than 1.
Now, the scale factor is given by
Since the scale factor is less than 1, so the dilation will be a reduction.
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Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
