This is a rather famous probability problem.
The easiest way to solve this is to calculate the probability that you WON'T roll a "double 6" (or a twelve) each time you roll the dice. There are 36 ways in which dice rols can appear and only one is a twelve. So, for one roll, the probability that you will NOT get a twelve is (35/36)^n where 35/36 is about .97222222 and n would equal 1 for the first trial. So for your first roll the odds that you WON'T get a 12 is .97222222.
For the second roll we calculate (35/36) to the second power or (35/36)^2 which equals about .945216.
When we get to the 24th roll we calculate (.97222222)^24 which equals 0.508596.
For the 25th roll, we calculate (.97222222)^25 which equals 0.494468. For the first time we have reached a probability which is lower than 50 per cent. That is to say, after 25 rolls, we have reached a point in which the probability is less than 50 per cent that we will NOT roll a twelve.
To phrase this more clearly, after 25 rolls we reach a point where the probability is greater then 50 per cent that you will roll a 12 at least once.
Please go to this page 1728.com/puzzle3.htm and look at puzzle 48. (The last puzzle on the page). An intersting story associated with this probability problem is that in 1952, a gambler named Fat the Butch bet someone $1,000 that he could roll a 12 after 21 throws. (He miscalculated the odds [as we know you need 25 throws] and after several HOURS, he lost $49,000!!!)
Please go that page and it has a link to the Fat the Butch story.
It would be 89.
You can also just look up a list of prime numbers online.
B. When the sky is cloudy it rains.
I hope this helps:)
If you add the price of the ticket and drink it comes out to 12.25, subtract from 25 and the answer is 12.75
Answer:
25.6 units
Step-by-step explanation:
From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).
First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:
where
are the coordinates of the first point
are the coordinates of the second point
- For AB:
![d=\sqrt{[1-(-5)]^{2}+(4-4)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5B1-%28-5%29%5D%5E%7B2%7D%2B%284-4%29%5E2%7D)
- For BC:
- For AC:
![d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5B3-%28-5%29%5D%5E%7B2%7D%20%2B%28-4-4%29%5E%7B2%7D%7D)
Next, now that we have our lengths, we can add them to find the perimeter of our triangle:
We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.
