Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as the histogram is not attached.
However, the solution to your question is to select the interval with the highest frequency
Take for instance, after getting readings from the histogram, yiu have:
1 - 10: 8
11 - 20: 5
21 - 30: 11
31 - 40: 2
And so on.......
Interval 11 - 20 will represent the required interval because it has the highest frequency.
To solve the quadratic equation given by 0=x^2-9x-20, we use the quadratic formula given by:
x=[-b+\- sqrt(b^2-4ac)]/(2a)
where,
a=1,b=-9,c=-20
thus substituting the above values into our formula we get:
x=[9+\-sqrt(9^2-4(-20*1))/(2*1)
x=[9+\-sqrt(161)]/2
x=[9+sqrt161]/2 or x=[9-sqrt161]/2
I think it is either a b or c
Answer:
8/π
Step-by-step explanation:
Volume of the cylinder is:
V = πr²h
Surface area of the cylinder (with square top and bottom) is:
A = 2πrh + 8r²
Use the volume equation to write h in terms of r.
h = V / (πr²)
Substitute into the area equation:
A = 2V / r + 8r²
Take derivative with respect to r.
dA/dr = -2V / r² + 16r
Set to 0 and solve for r.
0 = -2V / r² + 16r
2V / r² = 16r
2V = 16r³
V = 8r³
Plug into the volume equation.
8r³ = πr²h
8r = πh
h / r = 8 / π
Multiply 1 by 125, and 25 by x so you can solve for x.