Let X and Y be the digits.
If a two-digit number is written XY, then its value is 10x+y.
When the digits of the number are interchanged, it's YX, and its value is 10y+x.
The sum of the digits is 12.
The number formed by interchaning the digits is 54 more than the original number.
The digits X and Y are 3 and 9, so the original number is 39.
1.theoretical is not counting the results of the experimentssince there are an equal number of red and black and red, the probblity (theoretical) of picking a black one is 1/3
experimental=number of outcomes happened/total number of tests4 times black, 4 tests, so 4/4 or 100%=experimental proablity
2. experimental considered previous trials and theoretical did not
3. theoretical=(1/4) the trials don't influence each other so1/(4*4)=1/16
Answer:
1/6
1/12
1/2
Step-by-step explanation:
There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner. So there are a total of 3×4=12 possible combinations. We can show that by making a grid:
![\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26R%26B%26G%5C%5CR%26RR%26BR%26GR%5C%5CB%26BR%26BB%26BG%5C%5CP%26PR%26BP%26GP%5C%5CY%26RY%26BY%26GY%5Cend%7Barray%7D%5Cright%5D)
Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB). So the probability is 2/12 = 1/6.
There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.
There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP). So the probability is 6/12 = 1/2.
Answer:
12x+16
~Grouping~
12+16
4(3+4)
~Solution~
4(3+4)
Step-by-step explanation:
Hope that helps!!! :)
