Answer: The answer is provided below
Step-by-step explanation:
A histogram is a diagram which consist of rectangles whereby the area is proportional to frequency of a variable and the width is equal to class interval. A histogram is a commonly used graph that is used to show frequency distributions.
The cumulative histogram is a histogram whereby the vertical axis doesn't gives only the counts for a single bin, but gives the counts for that bin and all the bins for the maller values of a response variable.
Cumulative histograms are similar to normal histograms, but the main difference is that they graph cumulative frequencies unlike histograms that graph just frequencies.
Answer:
All you have to do is to the distributive method.
Step-by-step explanation:
(7x+14)(9x-8)
Answer:
- The shaded region is 9.83 cm²
Step-by-step explanation:
<em>Refer to attached diagram with added details.</em>
<h2>Given </h2>
Circle O with:
- OA = OB = OD - radius
- OC = OD = 2 cm
<h2>To find</h2>
<h2>Solution</h2>
Since r = OC + CD, the radius is 4 cm.
Consider right triangles OAC or OBC:
- They have one leg of 2 cm and hypotenuse of 4 cm, so the hypotenuse is twice the short leg.
Recall the property of 30°x60°x90° triangle:
- a : b : c = 1 : √3 : 2, where a- short leg, b- long leg, c- hypotenuse.
It means OC: OA = 1 : 2, so angles AOC and BOC are both 60° as adjacent to short legs.
In order to find the shaded area we need to find the area of sector OADB and subtract the area of triangle OAB.
Area of <u>sector:</u>
- A = π(θ/360)r², where θ- central angle,
- A = π*((mAOC + mBOC)/360)*r²,
- A = π*((60 + 60)/360))(4²) = 16.76 cm².
Area of<u> triangle AO</u>B:
- A = (1/2)*OC*(AC + BC), AC = BC = OC√3 according to the property of 30x60x90 triangle.
- A = (1/2)(2*2√3)*2 = 4√3 = 6.93 cm²
The shaded area is:
- A = 16.76 - 6.93 = 9.83 cm²
Input each points. (x,y) for each equation to find out if the system of equations are equivalent.
Point (-3,4) is a solution of this system of equations.
Point (-2,-6) isn't a solution of this system of equations.
Point (-4,3) is a solution of this system of equations.