Answer:
Part A: Option C
Part B: Option B
Step-by-step explanation:
Part A:
We have to find the equation of a straight line having x-intercept at (3,0) and y-intercept at (0,2).
Now, using intercept form of straight line equation we can write the equation of the required straight line is
⇒
⇒
So, option C is correct. (Answer)
Part B:
See the graph of the straight line provided.
The straight line passes through the points (10, 80) and (20, 100)
Therefore, the slope of the straight line is given by
therefore, option B is correct. (Answer)
Idk yet I did something like that yesterday
Answer:
It should be 1/16
Step-by-step explanation:
The word of in a word problem means multiplication so if you multiply you get 1/16
1507 are the different ways can 5 baseball players and 4 basketball players be selected from 12 baseball players and 13 basketball players
<em><u>Solution:</u></em>
Given that,
5 baseball players and 4 basketball players be selected from 12 baseball players and 13 basketball players
This is a combination problem
Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter
<em><u>The formula is given as:</u></em>
![n C_{r}=\frac{n !}{r !(n-r) !}](https://tex.z-dn.net/?f=n%20C_%7Br%7D%3D%5Cfrac%7Bn%20%21%7D%7Br%20%21%28n-r%29%20%21%7D)
Where n represents the total number of items, and r represents the number of items being chosen at a time
<em><u>Let us first calculate 5 baseball players from 12 baseball players</u></em>
Here, n = 12 and r = 5
![\begin{array}{l}{12 C_{5}=\frac{12 !}{5 !(12-5) !}} \\\\{12 C_{5}=\frac{12 !}{5 ! \times 7 !}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B12%20C_%7B5%7D%3D%5Cfrac%7B12%20%21%7D%7B5%20%21%2812-5%29%20%21%7D%7D%20%5C%5C%5C%5C%7B12%20C_%7B5%7D%3D%5Cfrac%7B12%20%21%7D%7B5%20%21%20%5Ctimes%207%20%21%7D%7D%5Cend%7Barray%7D)
<em><u>For a number n, the factorial of n can be written as:</u></em>
![n !=n \times(n-1) \times(n-2) \times \ldots . \times 2 \times 1](https://tex.z-dn.net/?f=n%20%21%3Dn%20%5Ctimes%28n-1%29%20%5Ctimes%28n-2%29%20%5Ctimes%20%5Cldots%20.%20%5Ctimes%202%20%5Ctimes%201)
Therefore,
![\begin{aligned}12 C_{5} &=\frac{12 \times 11 \times 10 \times \ldots \ldots \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\\\12 C_{5} &=\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2} \\\\12 C_{5} &=792\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D12%20C_%7B5%7D%20%26%3D%5Cfrac%7B12%20%5Ctimes%2011%20%5Ctimes%2010%20%5Ctimes%20%5Cldots%20%5Cldots%20%5Ctimes%202%20%5Ctimes%201%7D%7B5%20%5Ctimes%204%20%5Ctimes%203%20%5Ctimes%202%20%5Ctimes%201%20%5Ctimes%207%20%5Ctimes%206%20%5Ctimes%205%20%5Ctimes%204%20%5Ctimes%203%20%5Ctimes%202%20%5Ctimes%201%7D%20%5C%5C%5C%5C12%20C_%7B5%7D%20%26%3D%5Cfrac%7B12%20%5Ctimes%2011%20%5Ctimes%2010%20%5Ctimes%209%20%5Ctimes%208%7D%7B5%20%5Ctimes%204%20%5Ctimes%203%20%5Ctimes%202%7D%20%5C%5C%5C%5C12%20C_%7B5%7D%20%26%3D792%5Cend%7Baligned%7D)
<em><u>Similarly, 4 basketball players be selected 13 basketball players</u></em>
n = 13 and r = 4
Similarly we get,
![\begin{aligned}&13 C_{4}=\frac{13 !}{4 !(13-4) !}\\\\&13 C_{4}=\frac{13 !}{4 ! \times 9 !}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%2613%20C_%7B4%7D%3D%5Cfrac%7B13%20%21%7D%7B4%20%21%2813-4%29%20%21%7D%5C%5C%5C%5C%2613%20C_%7B4%7D%3D%5Cfrac%7B13%20%21%7D%7B4%20%21%20%5Ctimes%209%20%21%7D%5Cend%7Baligned%7D)
![13C_4 = 715](https://tex.z-dn.net/?f=13C_4%20%3D%20715)
<em><u>Thus total number of ways are:</u></em>
![12C_5 + 13C_4 = 792 + 715 = 1507](https://tex.z-dn.net/?f=12C_5%20%2B%2013C_4%20%3D%20792%20%2B%20715%20%3D%201507)
Thus there are 1507 different ways