The average rate of change (AROC) of a function f(x) on an interval [a, b] is equal to the slope of the secant line to the graph of f(x) that passes through (a, f(a)) and (b, f(b)), a.k.a. the difference quotient given by
![f_{\mathrm{AROC}[a,b]} = \dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5Ba%2Cb%5D%7D%20%3D%20%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
So for f(x) = x² on [1, 5], the AROC of f is
![f_{\mathrm{AROC}[1,5]} = \dfrac{5^2-1^2}{5-1} = \dfrac{24}4 = \boxed{6}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5B1%2C5%5D%7D%20%3D%20%5Cdfrac%7B5%5E2-1%5E2%7D%7B5-1%7D%20%3D%20%5Cdfrac%7B24%7D4%20%3D%20%5Cboxed%7B6%7D)
Answer: - 0.28
Explanation:
1) Expected value: is the weighted average of the values, being the probabilities the weight.
That is: ∑ of prbability of event i × value of event i.
In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).
2) Sample space:
Sum Points awarded
1+ 1 = 2 +6
1 + 2 = 3 +2
1 + 3 = 4 -1
1 + 4 = 5 -1
1 + 5 = 6 -1
1 + 6 = 7 -1
2 + 1 = 3 +2
2 + 2 = 4 -1
2 + 3 = 5 -1
2 + 4 = 6 -1
2 + 5 = 7 -1
2 + 6 = 8 -1
3 + 1 = 4 -1
3 + 2 = 5 -1
3 + 3 = 6 -1
3 + 4 = 7 -1
3 + 5 = 8 -1
3 + 6 = 9 -1
4 + 1 = 5 -1
4 + 2 = 6 -1
4 + 3 = 7 -1
4 + 4 = 8 -1
4 + 5 = 9 -1
4 + 6 = 10 -1
5 + 1 = 6 -1
5 + 2 = 7 -1
5 + 3 = 8 -1
5 + 4 = 9 -1
5 + 5 = 10 -1
5 + 6 = 11 +2
6 + 1 = 7 -1
6 + 2 = 8 -1
6 + 3 = 9 -1
6 + 4 = 10 -1
6 + 5 = 11 +2
6 + 6 = 12 +6
2) Probabilities
From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points
3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28
BC = √[(8-1)^2 +(1-6)^2]
BC = √(49+25
BC = √74
BC = 8.6
answer
8.6 units
Answer:
49
Step-by-step explanation:
100- 51 = 49
