Can Someone Please Help Me With This Problem.
1 answer:
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Answer:
a) f(x)
b) g(x)
c) for all values of x > 1, g(x) has a greater rate of change
Step-by-step explanation:
a) The rate of change of f(x) is the x-coefficient: 4. The average rate of change of g(x) on an interval can be found by dividing the change by the interval width. Here, the width of the interval of interest is 1-0 = 1, so the average rate of change is g(1) -g(0) = 4-2 = 2.
On the interval [0, 1], f(x) has the greater rate of change.
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b) The interval [2, 3] also has a width of 1, so the rate of change of g(x) on that interval is g(3) -g(2) = 20 -10 = 10. This value is greater than 4, so ...
on the interval [2, 3], g(x) has the greater rate of change.
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c) Part (b) demonstrates that g(x) will have a greater rate of change on some intervals. In fact, for any interval whose center is greater than x=1, g(x) will have a greater rate of change than f(x).
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In calculus terms, the rate of change of g(x) is g'(x) = 4x. This will be greater than 4 for any x > 1. The graph shows the rate of change is 4 at x=1, and is higher for x > 1. The average rate of change will be greater than 4 on any interval whose center is greater than x=1.
We can figure the average rate of change on the interval [a, b] as ...
m = (g(b) -g(a))/(b -a)
m = ((2b² +2) -(2a² +2))/(b -a) = 2(b² -a²)/(b-a) = 2(b+a)
For the average rate of change to exceed 4, the sum of the ends of the interval must exceed 2, which is to say the midpoint must exceed 1.
Answer:
The probability that the proportion of the core being used at any particular time will be less than 0.10 is 0.08146
Step-by-step explanation:
1/3 0.33 = 33.33 %
The Probability of that the proportion of the core being used at any particular time will be less than 0.10 is given by
PDF =
where x = 0.1
α = 2 and β = 4
PDF = = 1.458
CDF = = 0.08146
The probability that the proportion of the core being used at any particular time will be less than 0.10 = 0.08146.