Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The is the overall medium, is the medium of the first half, and is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between and . In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 + 
) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
Pressure
= Force/Area
Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa
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Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
