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
Answer:
The specific heat is 3.47222 J/kg°C.
Explanation:
Given that,
Temperature = 13°C
Temperature = 37°C
Mass = 60 Kg
Energy = 5000 J
We need to calculate the specific heat
Using formula of energy
Put the value into the formula
Hence, The specific heat is 3.47222 J/kg°C.
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
