Probabilty = 12×(1/12)^3=1/144
Answer:
Step-by-step explanation:
First you would see how many times 3 would go into 22 and it would be 7 with a remainder of 1. Then you would do 7 times 3 which is 21. And 22 minus 21 is 1. Then your answer would be 7 with a remainder of 1. If you don't want a remainder, and you want a decimal then you would add a decimal point after the 22 and then add a 0. You would bring down the zero and then the remainder of one would be 10 then you would see how much 3 goes into 10, and it only goes 3 times. Your answer would be 7.3 with a remainder of 1, If you keep going you will see it becomes a repeating decimal. *Hope it helped*
Answer:
Following the rule of BIDMAS
we will solve the one inside the bracket first, Addition will be solved before subtraction
i.e ( 5+4-2) x ( - 2)
= ( 9-2) x (-2)
= 7 x -2
= - 14
Step-by-step explanation:
A good place to start is to set 
to y. That would mean we are looking for 
to be an integer. Clearly, 
, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since 
is a radical, it only outputs values from ![[0,\infty]](https://tex.z-dn.net/?f=%20%5B0%2C%5Cinfty%5D%20)
, which means y is on the closed interval: ![[0,120]](https://tex.z-dn.net/?f=%20%5B0%2C120%5D%20)
.
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
to 
