The result of rolling a number cube 7 times is a 7-digit number composed of digits 1,2,3,4,5 and 6 so that digits can repeat. The total number of possibilities is 6^7.
The number of possibilities where 4 appears exactly two times is 5^5*(7!-6!/2).
5^5 is the number of 5-digits numbers composed of digits 1,2,3,5 and 6 so that digits can repeat.
7! is the number of permutations of digits 1,2,3,4,4,5 and 6.
6! is the number of permutations of digits 1,2,3,{4,4},5 and 6.
We don't want to subtract all numbers where digits 4 appear side by side. That's why we must divide 6! by 2.
Finally, the probability is P=5^5(7!-6!/2)/7^7
Answer:
D is the answer
Step-by-step explanation:
2/3 6/9 9/12 are three fractions with different numerator
Answer:
The first zero after decimal point and 4 only
Step-by-step explanation:
Despite having 5 decimal points, the rules of significant figures dictate that unless there is a digit other than zero after, the only significant numbers are those that come before zero. For this case, the significant digits are only 0.04 but if it was 0.0400005 then all the other zeros would have also be considered significant.
Answer:
what is the question I didn't get it please explain what we have to find
Step-by-step explanation:
