(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
Answer:
10.9
Step-by-step Explanation:
The Mean Absolute Deviation of a given data set tells us how far apart, on average, each data value is to the mean of the data set.
The smaller the Mean Absolute Deviation of a given data set is, the closer each data value is to the mean. This also implies less variability of the data set.
Invariably, the smaller the M.A.D, which connotes less variability, the more consistent the data set is.
Therefore, a M.A.D of 10.9 represents more consistency than a M.A.D of 15.2
22 i think idk sorry if i’m wrong
