Theres no such thing that means EHG for weight measurements
Answer:
x=3, y=5
Step-by-step explanation:
Answer:
If she wants at least one comedy, there are 1484 different combinations.
Step-by-step explanation:
The order in which she wants to pick the movies is not important. So, we use the combinations formula to solve this question.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
In this question:
She wants combinations of 3 movies, with at least one comedy. The easiest way to find this is finding the total number of combinations of 3 movies, from the set of 25(18 children's and 7 comedies), and subtract by the total number without comedies(which is 3 from a set of 25). So
Total:
3 from a set of 25.
![C_{25,3} = \frac{25!}{3!(25-3)!} = 2300](https://tex.z-dn.net/?f=C_%7B25%2C3%7D%20%3D%20%5Cfrac%7B25%21%7D%7B3%21%2825-3%29%21%7D%20%3D%202300)
Without comedies:
3 from a set of 18.
![C_{18,3} = \frac{18!}{3!(18-3)!} = 816](https://tex.z-dn.net/?f=C_%7B18%2C3%7D%20%3D%20%5Cfrac%7B18%21%7D%7B3%21%2818-3%29%21%7D%20%3D%20816)
At least one comedy:
![2300 - 816 = 1484](https://tex.z-dn.net/?f=2300%20-%20816%20%3D%201484)
If she wants at least one comedy, there are 1484 different combinations.
When managing money, you need to keep up a similar change proportion. Implying that 1 dollar separate 0.51 pound ought to equivalent x dollar partition 500 pound. So the condition will be 1/0.51 = x/500. Illuminate the condition by increasing both sides by 500 which makes x = 500/0.51
I hope this answer helps!
Answer:
k = 28
Step-by-step explanation:
![t = 6n - 1 \\ 167 = 6k - 1 \\ 6k = 168 \\ k = 28](https://tex.z-dn.net/?f=t%20%3D%206n%20-%201%20%5C%5C%20167%20%3D%206k%20-%201%20%5C%5C%206k%20%3D%20168%20%5C%5C%20k%20%3D%2028)