Me ! 1. an event manager recorded the number of people in different age groups that attended a music concert. a histogram titled concert audience is shown. the horizontal axis is labeled age group in years with bins 18 to 24, 25 to 31, 3 which data table accurately represents the data in the histogram? age group number of people 18–24 80 25–31 120 32–38 40 39–45 20 age group number of people 18–24 80 25–31 200 32–38 240 39–45 260 age group number of people 18–24 20 25–31 40 32–38 120 39–45 80 age group number of people 18–24 260 25–31 240 32–38 200 39–45 80 2. a dance school has 54 students who learn salsa, and 23 of those students also learn ballet. there are 15 students who do not learn salsa but learn ballet, and 10 students do not learn either salsa or ballet. which table best shows the relative frequency of rows for the data? (5 points) learn salsa do not learn salsa total learn ballet 0.29 0.19 1 do not learn ballet 0.39 0.13 1 total 0.68 0.32 1 learn salsa do not learn salsa total learn ballet 0.61 0.39 1 do not learn ballet 0.76 0.24 1 total 0.68 0.32 1 learn salsa do not learn salsa total learn ballet 0.43 0.60 1 do not learn ballet 0.57 0.4 1 total 0.68 0.32 1 learn salsa do not learn salsa total learn ballet 0.23 0.15 1 do not learn ballet 0.31 0.10 1 total 0.54 0.25 1

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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.