Glet x1, x2, . . , xn be a random sample from any distribution with mean µ and variance σ 2 . since σ 2 = v ar[x1] = e[(x − µ) 2 ], it makes sense to estimate this probability weighted average with a similar sample average such as σc2 : = xn i=1 (xi − x) 2 n . (a) show that this is a biased estimator of σ 2 . what is the bias? (b) adjust your estimator from part (a) to make it an unbiased estimator. your result will be known as the sample variance and will be denoted by s 2 . (note: some people use the more intuitive biased estimator given here as their definition of the sample variance and also denote it with s 2 . more people use the unbiased estimator you will be finding here. those are the cool people! )

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

answer: c

substitute in the values.

for a:

f(0) = 0 + 9

f(0) = 9

a is a solution.

for b:

f(7) = 14 + 9

f(7) = 23

b is a solution.

for c:

f (3) = 6 + 9

f(3) = 15

c is not a solution. it's supposed to be 14.

for d:

f(2) = 4 + 9

f(2)= 13

d is a solution.

the answer will be -6x^2 +46

step-by-step explanation: