By definition, the arc length is given by:
arc = (theta) * (R)
Where,
theta: central angle
R: radius
Substituting values we have:
arc = (π / 5) * (2.8)
Rewriting we have:
arc = ((3.14) / 5) * (2.8)
arc = 1.7584 cm
Round to the hundredth:
arc = 1.76 cm
Answer:
the arc length is:
arc = 1.76 cm
I am not sure about this but here goes nothing: d=2.5t+2.2 is very familiar to y=mx+b y=d m=2.5 x=t b=2.2 to find a new quasion we substitute d and t into place wich equals 1=2.5(0)+b simplification and you get b=1 so your answer is d=2.5t+1
Check the picture below.
so notice, their perimeter is the same, because the perimeter is just one rod anyway, and all rods are the same length, thus
Question not correct, so i have attached the correct question.
Answer:
SE = 0.59
Step-by-step explanation:
The mean of the students height is;
x' = (53 + 52.5 + 54 + 51 + 50.5 + 49.5 + 48 + 53 + 52 + 50)/10
x' = 51.35
Now, deviation from the mean for each height;
53 - 51.35 = 1.65
52.5 - 51.35 = 1.15
54 - 51.35 = 2.65
51 - 51.35 = -0.35
50.5 - 51.35 = -0.85
49.5 - 51.35 = -1.85
48 - 51.35 = -3.35
53 - 51.35 = 1.65
52 - 51.35 = 0.65
50 - 51.35 = -1.35
Now, square of the deviations above;
1.65² = 2.7225
1.15² = 1.3225
2.65² = 7.0225
-0.35² = 0.1225
-0.85² = 0.7225
-1.85² = 3.4225
-3.35² = 11.2225
1.65² = 2.7225
0.65² = 0.4225
-1.35² = 1.8225
Sum of the squared deviations;
2.7225 + 1.3225 + 7.0225 + 0.1225 + 0.7225 + 3.4225 + 11.2225 + 2.7225 + 0.4225 + 1.8225 = 31.525
Let's divide the sum by the DF of n - 1 i.e 10 - 1 = 9.
Thus;
31.525/9 = 3.50278
Taking the square root of that gives us the standard deviation.
Thus;
s = √3.50278
s = 1.8716
Formula for standard error is;
SE = s/√n
SE = 1.8716/√10
SE = 0.59