CPhill

P(drive l male )  =  .36/.75 = .48

P(drive) = .48

First box  ⇒    independent

Second box   ⇒    =

If the iameter of the circle is 6 inches, then  the radius is 3 in

So...the area of the circle is pi*(3)^2 = 9 pi in^2

The area of the board is  48 *24   = 1152 in^2

So.....the probabililty that the  circle  is hit is:

9pi/1152 ≈ 0.0245   =  2.45 %

Thanks, JGA

Sine wave for sure !!!  ⇒   "C"

Frequency =  1 / period  =   1 / 0.0038    =   263.16 Hz

Thanks, hectictar.....been missing you on here.......!!!!

A.  period

Thanks MediumRareRibeye....!!!!

Sounds like a good idea....!!!!!

I think this may prove it :

Let the sum of angles ABC and BCD = 240°

And let the bisectors of angles ABC and BCD  be as shown

Since AB = BC = CD

Then by SAS , triangle ABE  is congruent to triangle CBE

Similarly, by SAS, triangle CBE is congruent to triangle CDE

So....triangle CDE is congruent to triangle ABE is congruent to triangle CBE

So  angles DEC, AEB and BEC are congruent

But angles ABC and BCD sum to 240

So   1/2 ABC + 1/2 BCD = 120

But 1/2ABC = EBC

And 1/2 of BCD = BCE

So  angles EBC and BCE = 120

So angle BEC = 180 - (m EBC + BCE) = 180 - 120 = 60

But BEC = AEB = DEC.....so....

60 + 60 + 60  =  180

Therefore angle AED must be  straight a straight angle because  angle AEB + angle BEC + angle DEC  =  60 + 60 + 60  = 180

Thus....E must lie on AD.......and E is the intersection of the bisectors of angles ABC and BCD

Thanks, ILP!!!!