Answer:
37044 different combinations of 4 movies can he rent if he wants at least one comedy
Step-by-step explanation:
The order in which the movies are selected 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)
How many different combinations of 4 movies can he rent if he wants at least one comedy
The easier way to solve this is subtract the total from the number of combinations with no comedies.
Total:
4 movies from a set of 14 + 19 = 33. So
![C_{33,4} = \frac{33!}{4!(33-4)!} = 40920](https://tex.z-dn.net/?f=C_%7B33%2C4%7D%20%3D%20%5Cfrac%7B33%21%7D%7B4%21%2833-4%29%21%7D%20%3D%2040920)
No comedies:
4 movies from a set of 19.
![C_{19,4} = \frac{19!}{4!(19-4)!} = 3876](https://tex.z-dn.net/?f=C_%7B19%2C4%7D%20%3D%20%5Cfrac%7B19%21%7D%7B4%21%2819-4%29%21%7D%20%3D%203876)
At least one comedy:
40920 - 3876 = 37044
37044 different combinations of 4 movies can he rent if he wants at least one comedy
17 + 2 = 19 km is 1/2 of the total length, that is,
19 = (1/2)h
(1/2)h = 19 multiply both sides by 2
h = 2*19
h = 38 km
The total length in kilometers is 38 km
Answer:
7 hours
Explanation:
This week Melisa studied for 28 hours. Last week she studied 1/4'th of it.
So to find last week study hours. Divide this weeks study hours by 4.
Last week study hours:
- this week study hours ÷ 4
![\sf \rightarrow \dfrac{28}{4}](https://tex.z-dn.net/?f=%5Csf%20%5Crightarrow%20%5Cdfrac%7B28%7D%7B4%7D)
![\sf \rightarrow \dfrac{7}{1}](https://tex.z-dn.net/?f=%5Csf%20%5Crightarrow%20%5Cdfrac%7B7%7D%7B1%7D)
![\sf \rightarrow 7](https://tex.z-dn.net/?f=%5Csf%20%5Crightarrow%207)
Answer:
170 will have dogs
Step-by-step explanation:
Answer:
I hope your question is 4x²+3x+5, then the degree of this polynomial is 2