Here's a way to do it.
Let 4e +2 = 5n +1 . . . . . . for some integer n
Then e = (5n -1)/4 = n + (n -1)/4
We want (n-1)/4 to be an integer, so let it be integer m.
... m = (n -1)/4
... 4m = n -1
... 4m +1 = n
Substituting this into our expression for e gives
... e = (5(4m+1) -1)/4 = (20m +4)/4 = 5m +1
e = 5m+1 for any integer m
Answer:
C.1/6
Step-by-step explanation:
Initially the box has four $1 and six $5 bills. The probability of selecting a $5 bill in the first trial would be given as;
(number of $5 bills) / (total number of bills)
= (6)/(4+6) = 3/5
If in the first attempt we actually pick a $5 bill, the number of $5 bills will reduce by one to 5. Now, the probability of picking a $5 bill in the second attempt will be given as;
(new number of $5 bills) / (new total number of bills)
= (5)/(4+5) = 5/9
The new number of $5 bills will now be; 6 - 2 = 4 since we have already picked 2 without replacing them.
Now, the probability of picking a $5 bill in the third attempt will be given as;
(new number of $5 bills) / (new total number of bills)
= (4)/(4+4) = 1/2
Since the three attempts are independent, the probability of picking all three $5 bills is;
3/5 * 5/9 * 1/2 = 1/6
9514 1404 393
Answer:
-7/48
Step-by-step explanation:
Put the numbers in place of the corresponding variables and do the arithmetic.
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