Set an equation:
=44
divide both sides by 3.14
=14
divide both sides by 6.25, 4, and 3.14
=x/78.5
Since these two equations are equal, we can set them up that way.
x/78.5=14
multiply both sides by 78.5
x=1099
Answer: The maximum amount of water that container B can hold is 1,099 ounces.
Answer:
La arista de la bodega debe medir 6 metros.
Step-by-step explanation:
Podemos encontrar la arista de la bodega sabiendo su volumen:
En donde:
V: es el volumen de un cubo = 216 m³
a: es la arista del cubo =?
La arista del cubo es:
![a = \sqrt[3]{V} = \sqrt[3]{216 m^{3}} = 6 m](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7BV%7D%20%3D%20%5Csqrt%5B3%5D%7B216%20m%5E%7B3%7D%7D%20%3D%206%20m%20)
Por lo tanto, la arista de la bodega debe medir 6 metros.
Espero que te sea de utilidad!
Its an eighth 1/8. hope this helps you
I need help with my puzzle I can't figure it out j
Answer:
79% - almost 4 times out of 5.
The key to this is realizing that the number of games will not always be 5.
If A wins in a sweep - you have 2/3*(2/3)*2/3 percent chance of that happening - 8/27, or 29.63%
If A wins in 4, now we have 2/3*(2/3)*2/3*(1/3)*3 - the 1/3 is the chance that B wins a game. Note - there are only 3 ways B can win a game, not 4. B cannot win Game 4 because Game 4 would not be played in case of a sweep. That is why you cannot use a straight Pascal’s triangle to get your coefficients - the 1–4–6–4–1 is not possible if B cannot win Game 4. Anyway, the math is the same as the above, a 29.63% chance of A winning in 4.
For a 5 game set, A could lose 2 games in 6 possible ways (lose 1&2, 1&3, 1&4, 2&3, 2&4 or 3&4). Again, A cannot lose Game 5 - it would not be played once A wins 3 games. So the odds become 2/3*(2/3)*2/3*(1/3)*(1/3)*6, or 19.75%.
Add them up and you get 79.01%
Step-by-step explanation:
Hope it helps<3
