Cost= (5.50d+6.00p) +6.50
Coefficients: 5.50 and 6.00
Constant: 6.50
Variables: d and p
Answer:
Largest Median: Same
Largest Range: Castro
Largest IQR: Castro
Step-by-step explanation:
With a box-and-whisker plot, the box represents the upper and lower quartiles, the vertical line inside the box represents the median, and the lines on either side of the box show the high and low of the range.
Largest Median: Medians are the same because the verticle line inside the boxes is at 7 for both
Largest Range: Ms Castro's Class- the lines on either side of the box for Ms Castro go from 1-10 while the other class only goes from 4-10.
Largest IQR (interquartile range) Ms. Castro's class: their IQR goes from 5-8 while the other class only goes from 6-8
Here is your question: 7/6 divided by 2.

÷

.
Multiply and you get
Answer:
3x-10+70°=180°(Supplementary angles)
3x+60°=180°
3x=180°-60°
3x=120°
3x/3 =120°/3
x=40°
Step-by-step explanation:
Hope that this is helpful.
Have a wonderful day.
Answer:
Water needed for pool = 486 cubic feet
Plastic liner required for the pool = 298.3 feet
Step-by-step explanation:
Top view of the pool is a composite figure, having one rectangle and a trapezoid.
1). Water needed for the pool = volume of the pool
Volume of the pool = Area of the base × Depth
= (Area of the rectangle + Area of the trapezoid)× depth
Area of the trapezoid = 
= 
= 105 ft²
Area of the rectangle = Length × width
= 11 × 1.5
= 16.5 ft²
Now, volume of the pool = (105 + 16.5) × 4
= 121.5 × 4
= 486 cubic feet
b). Liner required = surface area of the pool excluding top
= Surface area of the walls + Area of the pool base
= (Perimeter of the pool) × depth + area of the base
= (12 + 11 + 1.5 + 10.7 + 9)×4 + 121.5
= 176.8 + 121.5
= 298.3 square feet
Therefore, amount of water required = 486 cubic feet
liner needed = 298.3 square feet