An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273-279), considered arthroscopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 13 of 18 repairs were successful while for shorter tears, 23 of 30 repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use alpha=0.05 . What is the P-value? (b) Calculate a one-sided 95% confidence bound on the difference in proportions that can be used to answer the question in part (a). (a) There is not evidence that the success rate is greater for longer tears. The P -value is Round your answer to four decimal places (e. G. 98.7654). (b) The one-sided 95% confidence bound is < p1-p2 . Round your answer to three decimal places (e. G. 98.765).

  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.

.95

step-by-step explanation: