CPhill

LOL!!!!  Nice one, ALAN !!!

{In my case, the digits would need to be followed by two more letters.....B.C. }

{Hey, jist ask zegroes......... }

The random point will be closer to 4 than it is to 0 when it occurs in this interval ......(2,5]

Since this represents about 3/5 of the portion of the number line from 0 to 5, the answer is 3/5 =0.6

Note that Melody's answer can be generalized as:

The measure of the exterior angle =

180 - [(n - 2)*180 / n]  =

[180n - 180n + 360] / n =

360/n   where n is the number of sides

Then.....  

360/n  = 20

20n = 360

n = 18

20mm

Actually.....I believe 91 will appear  4 times in the triangle.....

The entries are  C(14,2), C (14,12), C(91,1)  and C(91,90)

I think that's it  !!!!

Here's part of the question.......

a) The sums are

2 =  C(2,1)

6  = C(4, 2)

20  = C(6, 3)

70= C(8,4)

b) The pattern appears to be

[C(n,0)]^2  + [C(n, 1)]^2  + [C(n, 2)]^2 +.....+ [C(n,n)]^2   = C(2n, n)

Sorry....I don't know the answers to c or d   !!!!

Welcome, Heather.......!!!!!

Here's a way to do this with one variable only.

Let the son's age =x,  the daughter's age = 2x,   and the mother's age = 4x

And we have that in 22 years, the son wll be half as old as the mother......thus.....

2(x + 22)  = 4x + 22

2x + 44 = 4x + 22

22 = 2x

x = 11       and that's the son's age

2x = 22     and that's the daughter's age

log₅ (2x + 1)  = 1

In exponential form, we have

5¹ = 2x + 1

5 = 2x + 1     subtract 1 from both sides

4 = 2x      divide by 2 on both sides

x = 2

The total area of the dartboard  =  36pi

The "shaded" area between the circle with a radius of 4 and a radius of 2 = [4^2 - 2^2]pi = 12pi

So, the probability that any one dart thrown lands in this area = 12pi / 36pi  = 1/3

So, the probability that any dart lands in a "non-shaded" area = 1 - 1/3 = 2/3

So, if 9 darts are thrown, 2/3 of them should land in this area = 6 darts.

Here's a rough pic.....area "d" is the shaded one....areas "c" and "e" are the ones of interest.....