For 11, Figure B has the greatest perimeter. For 12, Figure C has an area of 108 square centimeters. For 13, Figure A has an area of 40 square feet.
Answer:
130=monthly charge
119=one time installation fee
Step-by-step explanation:
119 is the y-intercept and in problems like these, it is always the one time fee. 130 is the monthly payment because it has an x, which could account for the number of months. For example , say six months passed, you could multiply 130 * 6 to get the monthly fee for six months. I hope that makes sense!
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
![\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSavings%26Frequency%5C%5C%5C%240-%5C%24199%26339%5C%5C%5C%24200-%5C%24399%2686%5C%5C%5C%24400-%5C%24599%2655%5C%5C%5C%24600-%5C%24799%2618%5C%5C%5C%24800-%5C%24999%2611%5C%5C%5C%241000-%5C%241199%268%5C%5C%5C%241200-%5C%241399%263%5Cend%7Barray%7D%5Cright%5D)
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
First find out the function for the given table.
Find the slope.
m= (y2-y1)/(x2-x1)
m=(-1-(-3)) / (-1-(-2))
m=(-1+3) / (-1+2)
m=2/1
m=2
Now equation of line:
y-y1=m(x-x1)
y-(-3)=2(x-(-2))
y+3=2(x+2)
y+3=2x+4
y=2x+1
Now plot these points on the graph, we get the attached graph.
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
