CPhill

Here is  how guest got that answer  :

If all the digits are differerent  we have C(10,3)  = 120  sets

If  two  digits  are the  same  and  the  other  is different,  we  hve 10 ways to pick the two digits and 9 ways to select the remaining one  = 10 * 9  =  90 sets

If all 3 digits are the same,  we have 10  sets

So

120  +  90  +  10   =    220  sets  (ways)

A snack.....that sounds like a good idea   !!!!

Note  that   the   number  of possible different sandwiches  that  can be  made = 3 * 3 * 3   = 27

Since  he  is allergic  to   rye, tuna  and chipolte sauce,  the  the total possible sandwiches he  could  order  that  he is NOT allergic to  =   2 breads * 2meats  * 2 sauces   =   8

So.....the probability  that  he orders a sandwich  giving him an allergic reaction  =  1  -  8/27  =

19  / 27

P (1  or 3)   =   [ 8 + 9 ]  / 50  =    17  / 50

x  =  7 * 24 * 48  = 8064

Factor  24  =   3 * 2^3

Factor  48  =  3 *  2^4

So

x  =  7 * 3 * 3 * 2^3 * 2^4   =   7 * 3^2  * 2^7

Note  that for xy to be a perfect  cube, y  needs to be   7^2 * 3 * 2^2   =  588

So....to check

xy =  8064 * 588  =  4741362

Cube root    = 168

40%  * 4200  =  1680

If  this is shared  by  7   people, each get   1680  / 7   =  240 pounds =  amt  that Peter  gets in first scenerio

In the second scenerio, Peter   thinks  he will get   240 (1.25)   = 300 pounds

In  the second  scenerio, Peter   actually gets   4200 / 10    =  420  pounds

The equation  of the  line  is

y  = (2/3) ( x - - 18) - 12       simplify

y =  (2/3) ( x + 18)  - 12

y =  (2/3)x  + 12  - 12

y = (2/3)x 

To find the x intercept, let   y  = 0

0  = (2/3)x       multiply  both sides  by (3/2)

0  =  x       =   the x intercept

Don't feel bad.....I'm not sure  that anyone on here is as smart as they think they are.....LOL!!!!!

You're  on the right  track, Cal

1080   =  the sum of the interior angles

Assuming a regular polygon.....we have  

(n - 2) * 180   =  1080        divide both sides by 180

n - 2  =   6     add  2  to both sides

n  = 8   =    the number of sides

Good work, Cal   !!!

I believe   that you  are correct    !!!!!

That was a tough one......