Answer:
The dimensions of the work area are 16 ft by 36 ft
Step-by-step explanation:
Let
x ----> the length of the work area in feet
y-----> the width of the work area in feet
we know that
The perimeter of the work area is equal to
simplify
----> equation A
The area of the work area is equal to
---> equation B
substitute equation A in equation B
solve the quadratic equation by graphing
using a graphing tool
The solutions are
x=16 ft, y=36 ft
or
x=36 ft, y=16 ft
see the attached figure
therefore
The dimensions of the work area are 16 ft by 36 ft
Answer:
3.70 < 3.702 < 3 10/11 < 79/20
Step-by-step explanation:
9514 1404 393
Answer:
5/9 m/min
Step-by-step explanation:
The depth of the water is 2/5 of the depth of the trough, so the width of the surface will be 2/5 of the width of the trough:
2/5 × 2 m = 4/5 m
Then the surface area of the water is ...
A = LW = (18 m)(4/5 m) = 14.4 m²
The rate of change of height multiplied by the area gives the rate of change of volume:
8 m³/min = (14.4 m²)(h')
h' = (8 m³/min)/(14.4 m²) = 5/9 m/min
Answer:
The data is skewed to the bottom and contains an outlier.
Step-by-step explanation:
1. Test for outlier
An outlier is a point that is more than 1.5IQR below Q1 or above Q3.
IQR = Q3 - Q1 = 74 - 51 = 23
1.5 IQR = 1.5 × 23 = 34.5
51 - 15 = 36 > 1.5IQR
The point at 15 is an outlier.
2. Test for normal distribution
The median is not in the middle of the box.
Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.
3. Test for skewness
The longer part is to the left of the median, so the data is skewed left.
