<span>Maximum area = sqrt(3)/8
Let's first express the width of the triangle as a function of it's height.
If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have
w = 1 - 2b
b = h/sqrt(3)
So
w = 1 - 2*h/sqrt(3)
The area of the rectangle is
A = hw
A = h(1 - 2*h/sqrt(3))
A = h*1 - h*2*h/sqrt(3)
A = h - 2h^2/sqrt(3)
We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0.
We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3).
The midpoint is
(0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3)
So the desired height is 0.75/sqrt(3).
Now let's calculate the width:
w = 1 - 2*h/sqrt(3)
w = 1 - 2* 0.75/sqrt(3) /sqrt(3)
w = 1 - 2* 0.75/3
w = 1 - 1.5/3
w = 1 - 0.5
w = 0.5
The area is
A = hw
A = 0.75/sqrt(3) * 0.5
A = 0.375/sqrt(3)
Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens.
A = h - 2h^2/sqrt(3)
A' = 1h^0 - 4h/sqrt(3)
A' = 1 - 4h/sqrt(3)
Now solve for 0.
A' = 1 - 4h/sqrt(3)
0 = 1 - 4h/sqrt(3)
4h/sqrt(3) = 1
4h = sqrt(3)
h = sqrt(3)/4
w = 1 - 2*(sqrt(3)/4)/sqrt(3)
w = 1 - 2/4
w = 1 -1/2
w = 1/2
A = wh
A = 1/2 * sqrt(3)/4
A = sqrt(3)/8
And the other method got us 0.375/sqrt(3). Are they the same? Let's see.
0.375/sqrt(3)
Multiply top and bottom by sqrt(3)
0.375*sqrt(3)/3
Multiply top and bottom by 8
3*sqrt(3)/24
Divide top and bottom by 3
sqrt(3)/8
Yep, they're the same.
And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
Answer:
a) True
b) Parameter
Step-by-step explanation:
We are given the following in the question:
An accounting professor wants to know the average GPA of the students enrolled in her class.
Population:
- It is defined as the collection of all variables of interest.
- A sample of individuals of interest is drawn from a population.
For the given case
Individuals of interest:
Students enrolled in accounting class
Population of interest:
Students enrolled in accounting class
Characteristic of interest:
Average GPA of the students enrolled in her class.
a) The population is all students enrolled in the accounting class.
The given statement is true as it contains all the observation of all the individuals of interest.
b) The computed average GPA of all the students enrolled in the class is 3.29
Statistic is a descriptive measure that describes a sample where as a parameter is a measure that describes the population.
Since 3.29 is the average of all the students enrolled in her class that is the average GPA of population.
Thus, 3.29 is a parameter.
The domain is (-∞,∞)
the range is (−∞,2]
Answer:
Isosceles triangle
Step-by-step explanation:
Length AB
√(4 - 2)² + (7 - 2)²
√2² + 5²
√4 + 25
√29
Length AC
6 - 2 = 4
Length BC
√(6 - 4)² + (2 - 7)²
√2² + (-5)²
√4 + 25
√29
Length AB and BC is the same.
Length AC is the longest side.
Therefore, it is an isosceles triangle.
