Lin created a scaled copy of Triangle A with an area of 72 square units. How many times larger is the area of the scaled copy co
mpared to that of Triangle A
2 answers:
Answer:
The question is not complete, here is a possible match to the complete question:
Here is Triangle A. Lin created a scaled copy of Triangle A with an area of 72 square units. What scale factor did Lin apply to the Triangle A to create the copy? Remember: A=1/2bh
a) 4
b) 8
c) 16
Answer:
Scale factor = 16
Step-by-step explanation:
From the diagram attached to this solution, the triangle was plotted on a graph sheet, and each grid on the graph represents 1 unit. hence the dimensions of Triangle A from the diagram is as follows:
Base = 3 units
Height = 3 units
Next, in order to determine the scale factor of the area of the triangle after scaling, let us calculate the area of the unscaled triangle.
Area of Triangle = 1/2 (base × height)
Area or Triangle = 0.5 × 3 × 3 = 4.5 square units
Therefore,
Area of unscaled triangle = 4.5 squared units
Area of scaled triangle = 72 squared units
since the area of the scaled triangle is larger than the unscaled triangle, the scale factor is simply the number of times by which the scaled triangle was enlarged, compared to the unscaled triangle. This can be calculated by dividing the scaled triangle by the unscaled triangle as follows:
Scale factor =(scaled triangle) ÷ (unscaled triangle)
Scale factor = 72 ÷ 4.5 = 16
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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
