Answer:
1. mean is 17.4
2. interquartile range is 30
3. mean absolute deviation is 3
4. median is 51
Step-by-step explanation:
F(x) and g(x) were inverse functions, then if f(a) = b, then g(b) = a
f(x) = (5x+1) / x g(x) = x / (5x+1)
f(1) = 6, but g(6) = 6/31 ≠ 1
So, f(x) and g(x) are NOT inverse functions.
Answer: 70.909090909091%
Step-by-step explanation: Reading through the problem we have <em>78 is</em>, that's 78 equals, <em>what percent</em>, x/100, <em>of 110</em>, times 110.
It's important to understand that <em>percent</em> means over 100 so what percent would simply mean <em>x/100</em> or any variable but I will be using x.
So we have the equation 
So cross canceling the 110 and 100 to 11 and 10, we have 
Multiply both sides by 10 to get rid
of the fraction and we have 780 = 11x.
Now divide both sides by 11 and 70.909090909091 = x.
So, 78 is 70.909090909091% of 110.
Work is attached in the image provided.
Answer:
Verified below
Step-by-step explanation:
We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)
In trigonometric identities;
Cot θ = cos θ/sin θ
Thus;
(cot θ - 1)/(cot θ + 1) gives;
((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)
Simplifying numerator and denominator gives;
((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)
This reduces to;
>> (cos θ - sin θ)/(cos θ + sin θ)
Multiply top and bottom by ((cos θ + sin θ) to get;
>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)
In trigonometric identities, we know that;
cos 2θ = (cos² θ - sin²θ)
cos²θ + sin²θ = 1
sin 2θ = 2sinθcosθ
Thus;
(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:
>> cos 2θ/(1 + sin 2θ)
This is equal to the left hand side.
Thus, it is verified.
