Answer:
![\large \boxed{17}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B17%7D)
Step-by-step explanation:
Let's call the numbers "<em>tu</em>", where<em> t</em> is the tens digit and <em>u</em> is the units digit.
We can solve this by the "brute force" method — examining each number to see if it satisfies the conditions.
![\begin{array}{cccl}\mathbf{u}&\mathbf{tu}& \textbf{Divisible by u} & \\0 &20, 30, 40 & 0 & \text{Division by zero is impossible}\\1 & 11, 21, 31, 41 & 4& \text{All numbers are divisible by one}\\2 & 12, 22, 32, 42 & 4 & \text{All even numbers are divisible by two}\\3 & 13, 23, 33, 43 & 1 &\text{Only 33 is divisible by three}\\4 & 14, 24, 34, 44 & 2 & \text{Only 24 and 44 are divisible by four}\\5 & 15, 25, 35, 45 & 4 & \text{Numbers ending in 5 are divisible by five}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccl%7D%5Cmathbf%7Bu%7D%26%5Cmathbf%7Btu%7D%26%20%5Ctextbf%7BDivisible%20by%20u%7D%20%26%20%5C%5C0%20%2620%2C%2030%2C%2040%20%26%200%20%26%20%5Ctext%7BDivision%20by%20zero%20is%20impossible%7D%5C%5C1%20%26%2011%2C%2021%2C%2031%2C%2041%20%26%204%26%20%5Ctext%7BAll%20numbers%20are%20divisible%20by%20one%7D%5C%5C2%20%26%2012%2C%2022%2C%2032%2C%2042%20%26%204%20%26%20%5Ctext%7BAll%20even%20numbers%20are%20divisible%20by%20two%7D%5C%5C3%20%26%2013%2C%20%2023%2C%2033%2C%2043%20%26%201%20%26%5Ctext%7BOnly%2033%20is%20divisible%20by%20three%7D%5C%5C4%20%26%2014%2C%2024%2C%2034%2C%2044%20%26%202%20%26%20%5Ctext%7BOnly%2024%20and%2044%20are%20divisible%20by%20four%7D%5C%5C5%20%26%2015%2C%2025%2C%2035%2C%2045%20%26%204%20%26%20%5Ctext%7BNumbers%20ending%20in%205%20are%20divisible%20by%20five%7D%5C%5C%5Cend%7Barray%7D)
![\begin{array}{cccl}6 & 16, 26, 36, 46 & 1 & \text{Only 36 is divisible by six}\\7 & 17, 27, 37, 47 & 0 & \text{None of the numbers is divisible by seven}\\8 & 18, 28, 38, 48 & 1 & \text{Only 48 is divisible by eight}\\9 & 19, 29, 39, 49 & 0 & \text{None of the numbers is divisible by nine}\\& \textbf{TOTAL =} &\mathbf{17}& \\\end{array}\\\text{$\large \boxed{\mathbf{17}}$ numbers between 10 and 50 are divisible by their units digit}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccl%7D6%20%26%2016%2C%2026%2C%2036%2C%2046%20%26%201%20%26%20%5Ctext%7BOnly%2036%20is%20divisible%20by%20six%7D%5C%5C7%20%26%2017%2C%2027%2C%2037%2C%2047%20%26%200%20%26%20%5Ctext%7BNone%20of%20the%20numbers%20is%20divisible%20by%20seven%7D%5C%5C8%20%26%2018%2C%2028%2C%2038%2C%2048%20%26%201%20%26%20%5Ctext%7BOnly%2048%20is%20divisible%20by%20eight%7D%5C%5C9%20%26%2019%2C%2029%2C%2039%2C%2049%20%26%200%20%26%20%5Ctext%7BNone%20of%20the%20numbers%20is%20divisible%20by%20nine%7D%5C%5C%26%20%5Ctextbf%7BTOTAL%20%3D%7D%20%26%5Cmathbf%7B17%7D%26%20%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7B%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B17%7D%7D%24%20numbers%20between%2010%20and%2050%20are%20divisible%20by%20their%20units%20digit%7D)
We're going to be using combination since this question is asking how many different combinations of 10 people can be selected from a set of 23.
We would only use permutation if the order of the people in the committee mattered, which it seems it doesn't.
Formula for combination:
![C(n,r)=\dfrac{n!}{(n-r)!r!}](https://tex.z-dn.net/?f=C%28n%2Cr%29%3D%5Cdfrac%7Bn%21%7D%7B%28n-r%29%21r%21%7D)
Where
represents the number of objects/people in the set and
represents the number of objects/people being chosen from the set
There are 23 people in the set and 10 people being chosen from the set
![C(23,10)=\dfrac{23!}{(23-10)!10!}](https://tex.z-dn.net/?f=C%2823%2C10%29%3D%5Cdfrac%7B23%21%7D%7B%2823-10%29%2110%21%7D)
![=\dfrac{23!}{13!\times10!}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B23%21%7D%7B13%21%5Ctimes10%21%7D)
Usually I would prefer solving such fractions by hand instead of a calculator, but factorials can result in large numbers and there is too much multiplication. Using a calculator, we get
![=1,144,066](https://tex.z-dn.net/?f=%3D1%2C144%2C066)
Thus, there are 1,144,066 different 10 person committees that can be selected from a pool of 23 people. Let me know if you need any clarifications, thanks!
~ Padoru
Outcomes in a sample space :)
Answer:
A) 29/8
Step-by-step explanation:
On the left side of the equals sign, we have five x's and nine -1's.
On the right side of the equals sign, we have three -x's and twenty 1's.
Both sides are equal, so:
5(x) + 9(-1) = 3(-x) + 20(1)
5x − 9 = -3x + 20
Add 3x to both sides.
8x − 9 = 20
Add 9 to both sides.
8x = 29
Divide both sides by 8.
x = 29/8
3 feet because 36 divided by 12 is 3