A marketing firm is doing research for an Internet-based company. It wants to appeal to the age group of people who spend the most money online. The company wants to know if there is a difference in the mean amount of money people spend per month on Internet purchases depending on their age bracket. The marketing firm looked at two age groups, 18-24 years and 25-30 years, and collected the data shown in the following table. Let Population 1 be the amount of money spent per month on Internet purchases by people in the 18-24 age bracket and Population 2 be the amount of money spent per month on Internet purchases by people in the 25-30 age bracket. Assume that the population variances are not the same. Internet Spending per Month
18-24 Years25-30 Years
Mean Amount Spent56.8143.19
Standard Deviation16.9512.83
Sample Size2319
Step 1 of 2:
Construct a 95% confidence interval for the true difference between the mean amounts of money per month that people in these two age groups spend on Internet purchases. Round the endpoints of the interval to two decimal places, if necessary.
Interpret the confidence interval (0.29,14.27)(0.29,14.27) obtained in Step 1. and why?

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

your answer is b.

step-by-step explanation:

the 2 points are: (3,1) and (6,6)

first, let's find the slope:

(6-1)/(6-3) = 5/3

then let's plug this into the point-slope formula:

y - 1 = 5/3(x - 3)

this is your answer! the below is just the simplified form.

y - 1 = 5/3x - 5

y = 5/3x - 4