Class F=36.6666666667%
class E=33.3333333333%
class H=41.6666666667%
class G=<span>32%</span>
We have two points describing the diameter of a circumference, these are:

Recall that the equation for the standard form of a circle is:

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.
For this we use the following equation:

Now, we replace and solve:

The center of the circle is (-8,-7), so:

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

In this case, we will solve the delta with the center coordinate and the B coordinate.

Therefore, the equation for the standard form of a circle is:

In conclusion, the equation is the following:
Answer:
- <u><em>The height of the missing rectangle is 0.15</em></u>
Explanation:
The image attached has the mentioned <em>histogram</em>.
Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.
Only the rectangle for the class [3,4] is missing.
The height of each rectangle is the relative frequency of the corresponding class.
The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.
In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.
<u>1. Sum of the known relative frequencies</u>:
- 0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.85
<u>2. Missing frequency</u>:
<u>3. Conclusion</u>:
- The height of the missing rectangle is 0.15
Answer:
pppp
Step-by-step explanation:
ppppppppppppppppppppppppppp
sqrt (5x - 9) - 1 = x
sqrt(5x - 9) = x + 1
Square both sides:_
5x - 9 = x^2 + 2x + 1
x^2 - 3x + 10 = 0
(x - 5)(x + 2) = 0
x = 5 or -2.
Lets look for for any extraneous solutions:-
x = 5 ; sqrt (5*5-9) - 1 = sqrt16 - 1 = 3 and x = 3 (right hand side of equation)
so x = 5 is a solution
x = -2: sqrt(5*-2 - 9) - 1 = sqrt (-19) - 1 and x = -2 so this is extraneous
Answer:- One solution x = 5