Answer:
tHE ANSWER IS PIC
Step-by-step explanation:
Answer:
p^(a-1)
Step-by-step explanation:
The simplified expression can be seen in the picture attached below
As the factors have the same base, we just need to add or substract the exponents
p^(a-3+2) = p^(a-1)
Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
= ![10(9)(8)(7)\\=5040](https://tex.z-dn.net/?f=10%289%29%288%29%287%29%5C%5C%3D5040)
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
Answer:
The frequency of rolling a 3 or a 6 would be the same even if you try thousands or millions times.
The probability of rolling a 3 is: 1/ total outcomes probability = 1/6
The probability of rolling a 6 is: 1/ total outcomes probability = 1/6
The probability of rolling a 6 or a 3 = 2/ total outcomes probability = 2/6 = 1/3
Therefore, if you rolling the cube 600 times, the probability of rolling a 3 or a 6 is still 1/6. If 3 and 6 are both allowed, the probability would become 1/3
Hope this helped :3