I'm going to assume that your function is f(x) = 1 + x^2 (NOT x2).

I suspect you're trying to estimate the "area under the curve of f(x) = 1 + x^2. You need to use this or a similar description to explain what you're doing.

Also, you need to specify whether you want "left end points" or "right end points" or "midpoints." Again I must assume you want one or the other (and will assume that you meant "left end points").

First, let's address the case n=3. You must graph f(x) = 1 + x^2 between -1 and +1. We will find the "lower sum," using "left end points." The 3 x-values are {-1, -1/3, 1/3}. Evaluate the function f(x) = 1 + x^2 at these 3 x-values. Keep in mind that the interval width is 2/3.
The function (y) values are {0, 2/3, 4/3}.

Sorry, Michael, but I must stop here and await clarification from you regarding what you've been told to do in this problem. Otherwise too much guessing (regarding what you meant) is necessary. Please review the original problem and ensure that you have copied it exactly as presented, and also please verify whether this problem does indeed involve estimating areas under curves between starting and ending x-values.

7 0

2 years ago

The average estimated hours a person in the United States spent playing video games per year from 2002 to 2012 were 71, 80, 82,

ozzi

Answer:

Population variance 579

Population standard deviation 24

Step-by-step explanation:

The average estimated hours a person in the United States spent playing video games per year from 2002 to 2012 were 71, 80, 82, 78, 80, 91, 107, 121, 125, 131, and 142.

The population variance is given by:

$\sigma^2=\frac{\sum(x-\mu)^2}{n}$

$\mu=\frac{71+80+82+78+80+91+107+121+125+131+142}{11}$

$\mu=\frac{1108}{11}$

This implies that:

$\sigma^2=\frac{(71-\frac{1108}{11})^2+(80-\frac{1108}{11})^2+(82-\frac{1108}{11})^2+(78-\frac{1108}{11})^2+(80-\frac{1108}{11})^2+...+(142-\frac{1108}{11})^2}{11}$