On a test that has a normal distribution, a score of 60 falls two standard deviations above the mean, and a score of 42 falls on
2 answers:
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Refer to the figure shown below.
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²