Answer:
7.9 pounds of fruit
Step-by-step explanation:
In all means to add, so 7 + .9 = 7.9 pounds
Answer:
The third quartile is 56.45
Step-by-step explanation:
The given parameters are;
The first quartile, Q₁ = 30.8
The median or second quartile, Q₂ = 48.5
The mean,
= 42.0
Coefficient of skewness = -0.38
The Bowley's coefficient of skewness (SK) is given as follows;

Plugging in the values, we have;

Which gives;
-0.38×(Q₃ - 30.8) = Q₃ + 30.8 - 2 × 48.5
11.704 - 0.38·Q₃ = Q₃ - 66.2
1.38·Q₃ = 11.704 + 66.2 = 77.904
Q₃ = 56.45
The third quartile = 56.45.
The elimination method is a sufficient way to solve problems.
2x+y= 20
6x-5y=12
Add 5y to the one equation.
2x+6y= 20
6x= 12
Subtract 2x from both sides.
6y= 20
4x= 12
Divide 6 by 20.
y= 3.3
Divide 4 by 12.
x= 3
I hope this helped you!
Brainliest answer is appreciated!
Answer:
percentage of the total capacity is 75.6%
Step-by-step explanation:
Hello! To solve this problem we follow the following steps
1. draw the complete scheme of the problem (see attached image)
2. To solve this problem we must find the area of the circular sector using the following equation.(c in the second attached image)


3. observing the attached images we replace the values in the equations and find the area of the circular sector, remember that you must transform the angle to radians


4.we calculate the area of the total circle (At), then subtract the area of the circular sector (Ac) to find the area occupied by water (Aw)

Aw=At-Ac=153.93-37.57=116.36ft^2
5.Finally, we calculate the percentage that represents the water in the tank by dividing the area of the water over the total area of the tank

percentage of the total capacity is 75.6%
The answer is that it's a right triangle. One angle is 90°