Area of the sector = 1/2 * r^2 * theta where theta is the angle subtended by the arc at the center ( in radians)
In this case m < theta = 360 - 235 = 125 degrees or 2.182 radians
So, the area of shaded area = 1/2 * 20^2 * 2.182 = 436.4 in^2
This cannot be fit into a normal distribution. While the first 7 runners all have times relatively close, the time of the last runner is an outlier, as it is too far from the other data points. The standard deviation is calculated by taking the mean, subtracting each value from the mean, squaring the deviations, adding these, then dividing by the number of data points (7), and taking the square root. This gives an answer of SD = 0.1107.
A Because if you subtract his monthly salary and his amount paid then multiply the amount of 5% by the extra not excluding his monthly salary you get 6500$<span />
Answer:
If MN = XY, and XY = AB, then MN = AB. Transitive Property
If AB = CD, then CD = AB. Symmetric Property
Segment CD is congruent to segment CD. Reflexive Property
Step-by-step explanation:
Transitive property states that for all reall numbers a, b, and c, if a = b, b = c, then a = c.
Symmetric Property states that for all real numbers a and b, if a = b, then b = a.
Reflexive Property states that for every real number a, a=a
This is a problem of maxima and minima using derivative.
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =

Surface area of the rectangular sides =

Therefore, the total area of the cube is:

Isolating the variable y in terms of x:

Substituting this value in V:

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

Solving for x:

Solving for y:

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ftWidth = 4 ftHeight = 2 ftAnd its volume is: