Answer:
468.4 meters
Step-by-step explanation:
Find the width of the rectangle.
The area of a rectangle is A=length*width. Substitute the values of length and area and solve for width.
A=length*width
8,400=140*width
60= width
Use the width of the rectangle to find the circumference of the semicircles. Since each semicircle is half of a circle, the perimeter of the two semicircles is equal to the circumference of one circle.
The circumference of a circle is equal to pid, where d is the diameter. The diameter of the semicircle is the same as the width of the rectangle.
So, the diameter is 60 meters. Substitute the diameter into the formula for the circumference and simplify using 3.14 for pi.
≈188.4
(2 times 140)+ 188.4
So, the perimeter of the track is 468.4 meters.
Answer:
x<5
Step-by-step explanation:
View Picture
Answer:
2k³ is the GCF.
Step-by-step explanation:
GCF of 
Finding the factors:
=10k³[Factoring 10k³]
=2×5×k×k×k
=6k³[Factoring 6k³]
=2×3×k×k×k
Common factors in 10k³ and 6k³.
=2×k³
=2k³
Answer:
Mean and IQR
Step-by-step explanation:
The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.
For company A:
Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.
Hence, the mean will be the best measure of centre rather Than the median in this case.
For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.
