There are 120 seventh graders at Danielle's school. Each day, the students have the option of eating outside or in the lunch roo
m. Danielle decided to keep track of how many students eat outside each day. After recording her results for the whole school year, she has displayed the data for each month in a box-and-whisker plot. Below are her plots for the months of September and April.
-Fill in the blank-
The____number of students that ate outside in
mode, mean, median
April is approximately___
25, 10, 15 ,20
___Than the
greater, less
____number of students that ate outside in September.
median, mode, mean
There is more variability in the plot representing_____.
April, September
There is ____overlap in the two data sets.
much, some, no
-Fill in the blank-
The____number of students that ate outside in
mode, mean, median
April is approximately___
25, 10, 15 ,20
___Than the
greater, less
____number of students that ate outside in September.
median, mode, mean
There is more variability in the plot representing_____.
April, September
There is ____overlap in the two data sets.
much, some, no
2 answers:
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Complete Question:
ABCD Is a rectangle that represents a park.
The lines show all the paths in the park.
The circular path is in the centre of the rectangle and has a diameter of 15m.
Calculate the shortest distance from A to C across the park,using only the lines shown.
*See attachment for the diagram showing the paths in the park.
Answer:
Shortest distance from A to C through the paths only = 102.9 m
Step-by-step explanation:
Length of park = 80 m
Width of park = 50 m
Diameter of the circular path in the center of the park = 15 m
Therefore:
Shortest distance between from A to C across the park if we're to follow only the lines showing the oaths would be = (length of diagonal AC - diameter of the circular path) + (½ of the circumference of the circular path)
✔️Use pythagorean theorem to find AC:
AC = √(80² + 50²) = √8,900
AC = 94.3398113 ≈ 94.34 m (nearest hundredth)
✔️Diameter of circular path = 15 m
✔️½ of the circumference of the circular path = ½(πd)
d = 15 m
½ of the circumference = ½(π*15) = 23.5619449 ≈ 23.56 m (nearest hundredth)
✔️Shortest distance = (94.34 - 15) + 23.56 = 79.34 + 23.56 = 102.9 m
