Suppose i plan to drive across the san francisco bay bridge from berkeley, park at a parking facility near the san francisco airport (sfo), then take a parking shuttle from the parking facility to the airport departure terminal. there is a 30% chance that the bay bridge will be congested with traffic. if it is, it will take 1 hour to drive to the parking facility. if not, it will take 38 minutes to drive to the parking lot. the parking shuttle takes 10 minutes to get to the airport departure terminal from the parking lot. suppose it is equally likely that i must wait 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 minutes for the parking shuttle once i arrive at the parking lot, and that the amount of time i must wait for the parking shuttle is independent of the time it takes me to drive to the parking lot from berkeley. 1. the expected value of the time it takes to drive from berkeley to the airport parking lot . 2. the standard error of the time it takes to drive from berkeley to the airport parking lot minutes. 3. the expected value of the waiting time for a parking shuttle is . 4. the standard error of the waiting time for a parking shuttle . 5. the expected time it takes to get from berkeley to the san francisco airport by driving and taking the parking shuttle is . 6. the standard error of the time it takes to get from berkeley to the san francisco airport by driving and taking the parking shuttle .

c) yes, because of asa or aas

answer: y = 3/2 x - 3

y = 1/8 x - 3/4 that would be your correct option

step-by-step explanation: w

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement

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.