The accompanying data are lengths (inches) of bears. Find the percentile corresponding to 65.5 in.
1 answer:
Complete question is missing, so i have attached it.
Answer:
Percentile is 74th percentile
Step-by-step explanation:
All the lengths given are;
Bear Lengths 36.5 37.5 39.5 40.5 41.5 42.5 43.0 46.0 46.5 46.5 48.5 48.5 48.5 49.5 51.5 52.5 53.0 53.0 54.5 56.8 57.5 58.5 58.5 58.5 59.0 60.5 60.5 61.0 61.0 61.5 62.0 62.5 63.5 63.5 63.5 64.0 64.0 64.5 64.5 65.5 66.5 67.0 67.5 69.0 69.5 70.5 72.0 72.5 72.5 72.5 72.5 73.0 76.0 77.5
The number of lengths (inches) of bears given are 54 in number.
We are looking for the percentile corresponding to 65.5 in.
Looking at the lengths given, since they are already arranged from smallest to highest, let's locate the position of 65.5 in.
The position of 65.5 in is the 40th among 54 lengths given.
If the percentile is P, then;
P% x 54 = 40
P = (40 × 100)/54
P ≈ 74
You might be interested in
Check the picture below.
is not very specific above, but sounds like it's asking for an equation for the trapezoid only, mind you, there are square tiles too.
but let's do the trapezoid area then,
![\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad \sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20a%5E%7B%5Cfrac%7B%7B%20n%7D%7D%7B%7B%20m%7D%7D%7D%20%5Cimplies%20%20%5Csqrt%5B%7B%20m%7D%5D%7Ba%5E%7B%20n%7D%7D%20%5Cqquad%20%5Cqquad%0A%5Csqrt%5B%7B%20m%7D%5D%7Ba%5E%7B%20n%7D%7D%5Cimplies%20a%5E%7B%5Cfrac%7B%7B%20n%7D%7D%7B%7B%20m%7D%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
is not very specific above, but sounds like it's asking for an equation for the trapezoid only, mind you, there are square tiles too.
but let's do the trapezoid area then,
Answer:
H(t) = 15 -6sin(2.5π(t -0.5))
Step-by-step explanation:
For midline M, amplitude A, period T and time t0 at which the function is decreasing from the midline, the function can be written as ...
H(t) = M -Asin(2π/T(t -t0))
Using the given values of M=15, A=6, T=0.8 and t0 = 0.5, the equation is ...
H(t) = 15 -6sin(2.5π(t -0.5))
