To solve this problem you must apply the proccedure shown below:
1- You must apply the following formula:
Where
is the residual, 
is the observed value 
and 
is the predicted value .
2- You only need to substitute the 
into the equation 
and then, you must apply the formula for calculate the residual:
The answer is: The residual is 
F(x)=3x/2 for 0≤x≤2
<span>.....=6 - 3x/2 for 2<x≤4 </span>
<span>g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 </span>
<span>so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 </span>
<span>for x=1, h'(x)=-3/8 so h'(1)=-3/8 </span>
<span>When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 </span>
<span>When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 </span>
<span>h'(x)=3/8 so h'(3) = 3/8</span>
Answer:
22
Step-by-step explanation:
I would assume that the 6% rate would carry for the rest of the year, i.e. all 365 days, 6% of those have clouds interfering with stargazing
365 * 6% = 365 * 0.06 = 21.9, and round up to 22 because you can't really have .9 of a day here.
Answer:
b. Identify the variable of interest and state whether it is categorical or quantitative.
Step-by-step explanation:
Answer:
B. cos−1(StartFraction 11.9 Over 14.5 EndFraction) = θ
Step-by-step explanation:
From definition:
cos(θ) = adjacent/hypotenuse
The adjacent side respect angle GFE (or θ) is side FE, and side FG is the hypotenuse. Replacing with data and isolating θ:
cos(θ) = 11.9/14.5
θ = cos^-1(11.9/14.5)
