When you open a stepladder, you use a brace on each side of the ladder to lock the legs in place. Explain why the triangles form
ed on each side by the legs and the ground (AABC and ADEF in the diagram) are congruent.
2 answers:
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Answer:
Depth:
μ =20.2025 km
M = 15.625 km
Range = 47.15 km
σ ≈ 15.92 km
Q₁ = 5.7375 km
Q₃ = 34.6675 km
Magnitude:
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925
Step-by-step explanation:
The given data are;
Depth Magnitude
0.76 0.84
4.93 0.47
8.16 0.35
33.58 1.32
21.2 1.61
35.03 4.57
10.05 5.52
47.91 1.99
For the Depth, we have;
The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km
The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;
0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;
(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km
The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km
The Standard deviation of, σ, is given as follows;
Where;
= The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91
)
N = The total number of data point = 8
Substituting, (using Microsoft Excel) we get;
Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order
Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km
Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order
Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km
For the magnitude, we have, using the same formulas and procedures as above;
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925
Answer:
Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent an inverse variation if it can be expressed in the form or
In this problem we have
For ----->
so
Find the value of k
substitute the value of x and f(x)
therefore
the equation is equal to
For -------> substitute in the equation and find the value of f(x)