A shipment to a warehouse consists of 2,750 MP3 players. The manager chooses a random sample of 50
MP3 players and finds 6 that are defective. How many MP3 players in the shipment are likely to be
defective?

MP3 players in the shipment which are likely to be defective is 330.

Step-by-step explanation:

•Probability is event of an occurrence which is uncertain.

•Probability always lies between 0 to 1.

•The sample of 50 MP3 has 6 defective, to solve it has to be converted.

•Convert percentage in 100 multiply numerator and denominator by 2.

•Result is 6/50 converted to 12/100, bring it to 0.12%.

•12% is defected in 100% of sample.

•It is found here that 2750 is the population and 12% of 2750.

•2750 * 0.12 = 330, defective pieces in 2750.

•There are three types of Classical,Empirical or Experimental.

•Classical are ‘n’ number of events find the probability occurrence.  

•Empirical or experimental is purely based on events.

  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.