The first three are dependent variables as the outcomes could potentially change due to any of these circumstances.
The last one is an independent variable due to the fact that rolling a die several times will not affect the next outcome.
Answer:
the girls ate
of the box
Step-by-step explanation:
![\frac{3}{6} = \frac{1}{2} = \frac{2}{4} ,,,\frac{1}{4} +\frac{2}{4} =\frac{3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B6%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7B2%7D%7B4%7D%20%2C%2C%2C%5Cfrac%7B1%7D%7B4%7D%20%2B%5Cfrac%7B2%7D%7B4%7D%20%3D%5Cfrac%7B3%7D%7B4%7D)
Hope it can solve you problem... :)
You'd have no problem simplifying that expression if it said
-(5 cows) - (1 cow) + ( 14 cows) - (3 cows) .
You'd just add up the cows, and simplify the expression to say
5 cows .
When you see an expression that talks about 'x's instead of cows,
you get scared. It has something to do with algebra and math and
all that stuff, so you freeze.
Relax. Just make each 'x' a cow, and suddenly it almost solves itself.
Answer:
![{5}^{6} \: \: or \: \: {25}^{3} = 15625](https://tex.z-dn.net/?f=%20%7B5%7D%5E%7B6%7D%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%5C%3A%20%20%7B25%7D%5E%7B3%7D%20%20%3D%2015625)
Step-by-step explanation:
![{5}^{3} \times {5}^{3}](https://tex.z-dn.net/?f=%20%7B5%7D%5E%7B3%7D%20%20%5Ctimes%20%20%7B5%7D%5E%7B3%7D%20)
Here we can solve this in 2 ways ,
Method 1 :
![{5}^{3} \times {5}^{3}](https://tex.z-dn.net/?f=%20%7B5%7D%5E%7B3%7D%20%20%5Ctimes%20%20%7B5%7D%5E%7B3%7D%20)
Here the bases are equal. So,
![= > {5}^{3 + 3} = {5}^{6} = 15625](https://tex.z-dn.net/?f=%20%3D%20%20%3E%20%20%20%7B5%7D%5E%7B3%20%2B%203%7D%20%20%3D%20%20%7B5%7D%5E%7B6%7D%20%20%3D%2015625)
Method 2 :
![{5}^{3} \times {5}^{3}](https://tex.z-dn.net/?f=%20%7B5%7D%5E%7B3%7D%20%20%5Ctimes%20%7B5%7D%5E%7B3%7D)
Here the exponents are same. So,
![= > {(5 \times 5)}^{3} = {25}^{3} = 15625](https://tex.z-dn.net/?f=%20%3D%20%20%3E%20%20%7B%285%20%5Ctimes%205%29%7D%5E%7B3%7D%20%20%3D%20%20%7B25%7D%5E%7B3%7D%20%20%3D%2015625)