There is a data set consisting of the aluminum contents of 26 bihourly samples of recycled pet plastic from a recycling facility. those 26 measurements have latex: \frac{ }{y} y = 142.7 ppm and latex: s\approx98.2 s ≈ 98.2 ppm. use these facts to respond to the following. ( assume that n = 26 is large enough to permit the use of large-sample formulas in this case.) (a) make a 90% two-sided confidence interval for the mean aluminium content of such specimens over the 52- hour study period. (b) make a 95% two-sided confidence interval for the mean aluminium content of such specimens over the 52- hour study period. how does this compare to your answer to part (a)? (c) make a 90% upper confidence interval for the mean aluminium content of such specimens over the 52- hour study period. (find # such that (latex: -\propto − ∝ , #) is a 90% confidence interval.) how does this value compare to upper endpoint of your interval from part (a)? (d) make a 95% upper confidence interval for the mean aluminium content of such specimens over the 52- hour study period. how does this value compare to your answer to part (c)? (e) interpret your interval from (a) for someone with little statistical background. (speak in the context of the recycling study and use definition 2 as your guide.) html editor keyboard shortcuts

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

answer: 1475

step-by-step explanation:

becuase you have to know what they are asking you

thats one

step-by-step explanation:

inverse operation of addition