The relationship between what? Please give me a problem so I could help.
Answer:
![\frac{21}{32} = x](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B32%7D%20%3D%20x)
Step-by-step explanation:
![\frac{x}{\frac{3}{4}} = \frac{7}{8} \\ \\ \frac{3}{4} \times \frac{7}{8} = \frac{21}{32}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B%5Cfrac%7B3%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B7%7D%7B8%7D%20%5C%5C%20%5C%5C%20%5Cfrac%7B3%7D%7B4%7D%20%5Ctimes%20%5Cfrac%7B7%7D%7B8%7D%20%3D%20%5Cfrac%7B21%7D%7B32%7D)
Use multiplication [inverse of division] to figure this out.
I am joyous to assist you anytime.
Answer:
x=16
Step-by-step explanation:
Answer:
it is already in it's simple form
Answer:
a) 1,679,616 sequences
b) 6,561 sequences
c) 46,656 sequences
Step-by-step explanation:
a) How many different sequences are possible?
For each of the 8 dice, there are 6 possible outcomes. Therefore, the total number of difference sequences is:
![6^8=1,679,616\ sequences](https://tex.z-dn.net/?f=6%5E8%3D1%2C679%2C616%5C%20sequences)
b) How many different sequences consist entirely of even numbers?
For each of the 8 dice, there are 3 possible even outcomes. Therefore, the total number of difference sequences is:
![3^8=6,561\ sequences](https://tex.z-dn.net/?f=3%5E8%3D6%2C561%5C%20sequences)
c) How many different sequences are possible if the first, third, and fourth numbers must be the same?
If the first, third and fourth numbers are fixed, there are only 5 "free" positions, while there are 6 possible outcomes for the fixed positions since they have to be the same
![6*6^5 = 46,656\ sequences](https://tex.z-dn.net/?f=6%2A6%5E5%20%3D%2046%2C656%5C%20sequences)