Solution:
Number of times a die is rolled = 20
1 - 3=A
2 - 5=B
3 - 4=C
4 - 2=D
5 - 3=E
6 - 3=F
Total number of arrangements of outcomes , when a dice is rolled 20 times given that 1 appear 3 times, 2 appears 5 times, 3 appear 4 times, 4 appear 2 times , 5 appear three times, and 6 appear 3 times
= Arrangement of 6 numbers (A,B,C,D,E,F) in 6! ways and then arranging outcomes
= 6! × [ 3! × 5! × 4!×2!×3!×3!]
= 720 × 6×120×24×72→→[Keep in Mind →n!= n (n-1)(n-2)(n-3)........1]
= 895795200 Ways
Answer:
a =1 and a=4.
Step-by-step explanation:
The function is
If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.
Now, for a a positive real we have that 
will annulate the denominator, i.e
. But, if a = 1 we have:
so, the value 
won't annulate the denominator.
Now, for a = 4 we have:
so, the value 
won't annulate the denominator.
In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.
The top half is called the numerator.
The bottom is called the denominator.
Yeah the answer is 35 because if you multiply 5x7 it will give u 35 for an answer
