CPhill

33.   P ( First day is science  l  Female  )  =   80  / (80 + 80 + 70)  =  80  / 230   =  8/23  = 0.348

34.       81, 85, 82, 93, 85, 84, 95, 87, 88, 91

Sort the data from low to high

81, 82, 84, 85, 85, 87, 88, 91, 93, 95

Mean   =   sum  of the scores/ number of scores  =  87.1

Median  = (85 + 87) / 2  = 86

Mode = 85

THX, Melvsky....!!!!!

The events are independent....thus...the probability of selecting a Blazer on the second draw  is the same as the first draw....  80 / 400   =  1/5

Last one

Angle S intercepts  minor arc RT....its measure   = 1/2 of this arc  = 1/2 * 60°  =  30°

Angle R intercepts  minor arc QS ...the measure of minor arc  QS  is twice the measure  of angle R  =2 * 37  = 74°

For the measure of RQ....draw PQ....since PQ  = PR, then in triangle PQR.....the angles opposite these sides are equal

So angle  QRP  = angle RQP  = 37°

So...angle RPQ  = 180  - 37  - 37  = 180 - 74  =   106°

And angle RPQ is a central angle  subtending arc RQ....so....RQ  has the same measure  = 106° 

39....

If the mean  is 64  and the standard deviation is 4, 68%  of the bags will fall  between  60 - 68

And 1/2 of these will  fall between  64-68....so    (1/2) * 68%  = 34%

26

We can apply the secant-tangent theorem, here....since AB  meets radius AP.....then AB  is tangent to the circle at A

And we have this

AB^2  = PB * CB

12^2  = PB  * 8

144  = PB  * 8      divide both sides by  8

18  = PB

So...PB  is the  hypotenuse of  triangle APB.....and by the Pythagorean Theorem, 

PB   = √[ AP^2  + AB^2]       square both sides

PB^2  = AP^2  + AB^2

324 = 144 + AB^2     subtract 144 from both sides

180  = AB^2       take the square root of both sides

√180  =  AB   =  2√45  = AB  =  "x"

20....No...the measure of angle ADC  would have to be equal to the measure of angle BDC for the two arcs to  be equal....but....we don't know their measures

24.... Two tangents drawn to a circle  from a point are  equal in length

Therefore....BC  = AB  = 3

And CD  = ED  = 4

So...x  = BC + CD  = 3 + 4  = 7

OK...post away.......we'll see if we know any of the answers.....!!!!

19....

Since  minor arc MN  = minor arc  QP......then the chords span  equal arcs......and, by Euclid, chords spanning equal arcs  are equidistant from the center of a circle

15.

Since the  diameter is perpendicular to the chord AB,  then it bisescts  AB....thus.....AE  = BE  =  6

And the intersecting  chord theorem says this

DE * CE  = AE  * BE

DE  *  2  =   6 * 6

DE  * 2   = 36     divide both sides by 2

DE  = 18

So....the diameter  =  DE + CE  = 18  + 2   =  20

And the radius is  1/2 of this  = 10